PHP Echo Appearing Twice

-1

I am trying to show an echo when a user improperly enters their license key, though, for some reason, the message appears twice.

    $invalidkey = '<!DOCTYPE HTML>

            <html>

                <body>

                    <center>

                    <div class="container2"

                        <h1>Your Product Key is Invalid!</h1>

                    </div>

                    </center>

                </body>

            </html>

            ';



if ($resulto->num_rows > 0) 

{

    while($row = $resulto->fetch_assoc()) 

    {

        $user_group = $row["LicenseKey"];

        $days = $row["Count"];

        // Key is valid

        if ($user_group == $key)

        {

            $keyvalidated = true;

            echo $user_group;

            echo $key;



        }

        // Key is invalid

        else if ($usergroup !== $key)

        {

            echo $invalidkey;

            $keyvalidated = false;

        }    

    }

}

Here is an image of the error actually appearing:

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

asked Nov 12 '18 at 5:16

Hytax

2

perhaps you have 2 results, so 2 lopps,
– user10226920
Nov 12 '18 at 5:20

1

Because it is inside the loop, then it runs twice... change your while loop condition
– Vpa
Nov 12 '18 at 5:21

I wouldn't echo an entire HTML document into what I can only assume is already a partial HTML document. That's going to cause you some rendering problems even after you solve this one.
– Phil
Nov 12 '18 at 5:22

What is $usergroup and what is $key. How many rows are in your result-set?
– Phil
Nov 12 '18 at 5:24

add a comment |

-1

I am trying to show an echo when a user improperly enters their license key, though, for some reason, the message appears twice.

    $invalidkey = '<!DOCTYPE HTML>

            <html>

                <body>

                    <center>

                    <div class="container2"

                        <h1>Your Product Key is Invalid!</h1>

                    </div>

                    </center>

                </body>

            </html>

            ';



if ($resulto->num_rows > 0) 

{

    while($row = $resulto->fetch_assoc()) 

    {

        $user_group = $row["LicenseKey"];

        $days = $row["Count"];

        // Key is valid

        if ($user_group == $key)

        {

            $keyvalidated = true;

            echo $user_group;

            echo $key;



        }

        // Key is invalid

        else if ($usergroup !== $key)

        {

            echo $invalidkey;

            $keyvalidated = false;

        }    

    }

}

Here is an image of the error actually appearing:

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

asked Nov 12 '18 at 5:16

Hytax

2

perhaps you have 2 results, so 2 lopps,
– user10226920
Nov 12 '18 at 5:20

1

Because it is inside the loop, then it runs twice... change your while loop condition
– Vpa
Nov 12 '18 at 5:21

I wouldn't echo an entire HTML document into what I can only assume is already a partial HTML document. That's going to cause you some rendering problems even after you solve this one.
– Phil
Nov 12 '18 at 5:22

What is $usergroup and what is $key. How many rows are in your result-set?
– Phil
Nov 12 '18 at 5:24

add a comment |

-1

I am trying to show an echo when a user improperly enters their license key, though, for some reason, the message appears twice.

    $invalidkey = '<!DOCTYPE HTML>

            <html>

                <body>

                    <center>

                    <div class="container2"

                        <h1>Your Product Key is Invalid!</h1>

                    </div>

                    </center>

                </body>

            </html>

            ';



if ($resulto->num_rows > 0) 

{

    while($row = $resulto->fetch_assoc()) 

    {

        $user_group = $row["LicenseKey"];

        $days = $row["Count"];

        // Key is valid

        if ($user_group == $key)

        {

            $keyvalidated = true;

            echo $user_group;

            echo $key;



        }

        // Key is invalid

        else if ($usergroup !== $key)

        {

            echo $invalidkey;

            $keyvalidated = false;

        }    

    }

}

Here is an image of the error actually appearing:

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

asked Nov 12 '18 at 5:16

Hytax

I am trying to show an echo when a user improperly enters their license key, though, for some reason, the message appears twice.

    $invalidkey = '<!DOCTYPE HTML>

            <html>

                <body>

                    <center>

                    <div class="container2"

                        <h1>Your Product Key is Invalid!</h1>

                    </div>

                    </center>

                </body>

            </html>

            ';



if ($resulto->num_rows > 0) 

{

    while($row = $resulto->fetch_assoc()) 

    {

        $user_group = $row["LicenseKey"];

        $days = $row["Count"];

        // Key is valid

        if ($user_group == $key)

        {

            $keyvalidated = true;

            echo $user_group;

            echo $key;



        }

        // Key is invalid

        else if ($usergroup !== $key)

        {

            echo $invalidkey;

            $keyvalidated = false;

        }    

    }

}

Here is an image of the error actually appearing:

php echo

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

asked Nov 12 '18 at 5:16

Hytax

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

asked Nov 12 '18 at 5:16

Hytax

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

edited Nov 12 '18 at 5:53

SherylHohman

4,56063246

asked Nov 12 '18 at 5:16

Hytax

asked Nov 12 '18 at 5:16

Hytax

asked Nov 12 '18 at 5:16

Hytax

2

perhaps you have 2 results, so 2 lopps,
– user10226920
Nov 12 '18 at 5:20

1

Because it is inside the loop, then it runs twice... change your while loop condition
– Vpa
Nov 12 '18 at 5:21

I wouldn't echo an entire HTML document into what I can only assume is already a partial HTML document. That's going to cause you some rendering problems even after you solve this one.
– Phil
Nov 12 '18 at 5:22

What is $usergroup and what is $key. How many rows are in your result-set?
– Phil
Nov 12 '18 at 5:24

add a comment |

2

perhaps you have 2 results, so 2 lopps,
– user10226920
Nov 12 '18 at 5:20

1

Because it is inside the loop, then it runs twice... change your while loop condition
– Vpa
Nov 12 '18 at 5:21

I wouldn't echo an entire HTML document into what I can only assume is already a partial HTML document. That's going to cause you some rendering problems even after you solve this one.
– Phil
Nov 12 '18 at 5:22

What is $usergroup and what is $key. How many rows are in your result-set?
– Phil
Nov 12 '18 at 5:24

perhaps you have 2 results, so 2 lopps,
– user10226920
Nov 12 '18 at 5:20

Because it is inside the loop, then it runs twice... change your while loop condition
– Vpa
Nov 12 '18 at 5:21

I wouldn't echo an entire HTML document into what I can only assume is already a partial HTML document. That's going to cause you some rendering problems even after you solve this one.
– Phil
Nov 12 '18 at 5:22

What is $usergroup and what is $key. How many rows are in your result-set?
– Phil
Nov 12 '18 at 5:24

add a comment |

3 Answers
3

active

oldest

votes

My guess is that for whatever reason there are actually two records in the result set. The best fix would be to find a way to run the right query, which just returns a single record for a single user. As a quick fix, perhaps just check the first record:

if ($resulto->num_rows > 0) {

    $row = $resulto->fetch_assoc();

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key) {

        $keyvalidated = true;

        echo $user_group;

        echo $key;

    }

    // Key is invalid

    else if ($usergroup !== $key) {

        echo $invalidkey;

        $keyvalidated = false;

    }    

}

Again, you need to find out why the result set has two, or more than one, records.

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

The quick fix might mean that unexpected results occur, because the wrong record is always selected and the user's data isn't properly validated.
– symlink
Nov 12 '18 at 5:54

add a comment |

Please try adding a break since there could be more than one iteration (or you wouldn't need a loop):

// Key is invalid

else if ($usergroup !== $key)

{

    echo $invalidkey;

    $keyvalidated = false;

    break;

}

answered Nov 12 '18 at 5:21

bestinamir

5917

i love you sir <3
– Hytax
Nov 12 '18 at 5:24

add a comment |

The reason why it's showing 2 times is because query is fetching 2 results. So you need to limit query to just 1, just read the first record, or break the loop after 1st iteration.

while($row = $resulto->fetch_assoc()) 

{

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key)

    {

        $keyvalidated = true;

        echo $user_group;

        echo $key;



    }

    // Key is invalid

    else if ($usergroup !== $key)

    {

        echo $invalidkey;

        $keyvalidated = false;

    }    



    break;

}

answered Nov 12 '18 at 5:20

M A

35829

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53256294%2fphp-echo-appearing-twice%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

if ($resulto->num_rows > 0) {

    $row = $resulto->fetch_assoc();

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key) {

        $keyvalidated = true;

        echo $user_group;

        echo $key;

    }

    // Key is invalid

    else if ($usergroup !== $key) {

        echo $invalidkey;

        $keyvalidated = false;

    }    

}

Again, you need to find out why the result set has two, or more than one, records.

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

The quick fix might mean that unexpected results occur, because the wrong record is always selected and the user's data isn't properly validated.
– symlink
Nov 12 '18 at 5:54

add a comment |

if ($resulto->num_rows > 0) {

    $row = $resulto->fetch_assoc();

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key) {

        $keyvalidated = true;

        echo $user_group;

        echo $key;

    }

    // Key is invalid

    else if ($usergroup !== $key) {

        echo $invalidkey;

        $keyvalidated = false;

    }    

}

Again, you need to find out why the result set has two, or more than one, records.

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

The quick fix might mean that unexpected results occur, because the wrong record is always selected and the user's data isn't properly validated.
– symlink
Nov 12 '18 at 5:54

add a comment |

if ($resulto->num_rows > 0) {

    $row = $resulto->fetch_assoc();

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key) {

        $keyvalidated = true;

        echo $user_group;

        echo $key;

    }

    // Key is invalid

    else if ($usergroup !== $key) {

        echo $invalidkey;

        $keyvalidated = false;

    }    

}

Again, you need to find out why the result set has two, or more than one, records.

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

if ($resulto->num_rows > 0) {

    $row = $resulto->fetch_assoc();

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key) {

        $keyvalidated = true;

        echo $user_group;

        echo $key;

    }

    // Key is invalid

    else if ($usergroup !== $key) {

        echo $invalidkey;

        $keyvalidated = false;

    }    

}

Again, you need to find out why the result set has two, or more than one, records.

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

answered Nov 12 '18 at 5:21

Tim Biegeleisen

218k1387140

The quick fix might mean that unexpected results occur, because the wrong record is always selected and the user's data isn't properly validated.
– symlink
Nov 12 '18 at 5:54

add a comment |

The quick fix might mean that unexpected results occur, because the wrong record is always selected and the user's data isn't properly validated.
– symlink
Nov 12 '18 at 5:54

The quick fix might mean that unexpected results occur, because the wrong record is always selected and the user's data isn't properly validated.
– symlink
Nov 12 '18 at 5:54

add a comment |

Please try adding a break since there could be more than one iteration (or you wouldn't need a loop):

// Key is invalid

else if ($usergroup !== $key)

{

    echo $invalidkey;

    $keyvalidated = false;

    break;

}

answered Nov 12 '18 at 5:21

bestinamir

5917

i love you sir <3
– Hytax
Nov 12 '18 at 5:24

add a comment |

Please try adding a break since there could be more than one iteration (or you wouldn't need a loop):

// Key is invalid

else if ($usergroup !== $key)

{

    echo $invalidkey;

    $keyvalidated = false;

    break;

}

answered Nov 12 '18 at 5:21

bestinamir

5917

i love you sir <3
– Hytax
Nov 12 '18 at 5:24

add a comment |

Please try adding a break since there could be more than one iteration (or you wouldn't need a loop):

// Key is invalid

else if ($usergroup !== $key)

{

    echo $invalidkey;

    $keyvalidated = false;

    break;

}

answered Nov 12 '18 at 5:21

bestinamir

5917

Please try adding a break since there could be more than one iteration (or you wouldn't need a loop):

// Key is invalid

else if ($usergroup !== $key)

{

    echo $invalidkey;

    $keyvalidated = false;

    break;

}

answered Nov 12 '18 at 5:21

bestinamir

5917

answered Nov 12 '18 at 5:21

bestinamir

5917

answered Nov 12 '18 at 5:21

bestinamir

5917

answered Nov 12 '18 at 5:21

bestinamir

5917

i love you sir <3
– Hytax
Nov 12 '18 at 5:24

add a comment |

i love you sir <3
– Hytax
Nov 12 '18 at 5:24

i love you sir <3
– Hytax
Nov 12 '18 at 5:24

add a comment |

The reason why it's showing 2 times is because query is fetching 2 results. So you need to limit query to just 1, just read the first record, or break the loop after 1st iteration.

while($row = $resulto->fetch_assoc()) 

{

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key)

    {

        $keyvalidated = true;

        echo $user_group;

        echo $key;



    }

    // Key is invalid

    else if ($usergroup !== $key)

    {

        echo $invalidkey;

        $keyvalidated = false;

    }    



    break;

}

answered Nov 12 '18 at 5:20

M A

35829

add a comment |

The reason why it's showing 2 times is because query is fetching 2 results. So you need to limit query to just 1, just read the first record, or break the loop after 1st iteration.

while($row = $resulto->fetch_assoc()) 

{

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key)

    {

        $keyvalidated = true;

        echo $user_group;

        echo $key;



    }

    // Key is invalid

    else if ($usergroup !== $key)

    {

        echo $invalidkey;

        $keyvalidated = false;

    }    



    break;

}

answered Nov 12 '18 at 5:20

M A

35829

add a comment |

The reason why it's showing 2 times is because query is fetching 2 results. So you need to limit query to just 1, just read the first record, or break the loop after 1st iteration.

while($row = $resulto->fetch_assoc()) 

{

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key)

    {

        $keyvalidated = true;

        echo $user_group;

        echo $key;



    }

    // Key is invalid

    else if ($usergroup !== $key)

    {

        echo $invalidkey;

        $keyvalidated = false;

    }    



    break;

}

answered Nov 12 '18 at 5:20

M A

35829

The reason why it's showing 2 times is because query is fetching 2 results. So you need to limit query to just 1, just read the first record, or break the loop after 1st iteration.

while($row = $resulto->fetch_assoc()) 

{

    $user_group = $row["LicenseKey"];

    $days = $row["Count"];

    // Key is valid

    if ($user_group == $key)

    {

        $keyvalidated = true;

        echo $user_group;

        echo $key;



    }

    // Key is invalid

    else if ($usergroup !== $key)

    {

        echo $invalidkey;

        $keyvalidated = false;

    }    



    break;

}

answered Nov 12 '18 at 5:20

M A

35829

answered Nov 12 '18 at 5:20

M A

35829

answered Nov 12 '18 at 5:20

M A

35829

answered Nov 12 '18 at 5:20

M A

35829

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Nrthugu