Form not displaying in PHP

Question

So I have a page using PHP and a MySQL query. What I'm wanting to do is create basically an "edit" page that takes data from my database and uses it to show the values in various inputs. The user can then change the data in the input which will then update the corresponding MySQL table row. However, for whatever reason the page is NOT displaying the form, but rolling over to the else statement. I can verify the $_SESSION['weaponName'] is working, because it will echo the correct thing. Any ideas on why the form will not show up for me?

edit.php

   <?php
session_start();

$con=mysqli_connect("localhost","username","password","db_name");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$weaponName = $_SESSION['weaponName'];

$query = mysqli_query($con, "SELECT * FROM weapons limit 1");

if(mysqli_num_rows($query)>=1){
while($row = mysqli_fetch_array($query)) {
    $creator= $row['creator'];
    $weaponCategory= $row['weaponCategory'];
    $weaponSubCategory= $row['weaponSubCategory'];
    $costAmount= $row['costAmount'];
    $costType= $row['costType'];
    $damageS= $row['damageS'];
    $damageM= $row['damageM'];
    $critical= $row['critical'];
    $rangeIncrement= $row['rangeIncrement'];
    $weight= $row['weight'];
    $weaponType= $row['weaponType'];
    $masterwork= $row['masterwork'];
    $attributes= $row['attributes'];
    $specialAbilities= $row['specialAbilities'];
    $additionalInfo= $row['additionalInfo'];
}

?>

<form action="weaponEditUpdate.php" method="post">
    <input type="hidden" name="weaponName" value="<?php echo $weaponName;?>">

    Weapon Name: <input type="text" name="weaponName" value="<?php echo $weaponName;?>">
    <br>
    Weapon Category: <select name="weaponCategory">
    <?php while ($row = mysqli_fetch_array($query)) {
    echo "<option value='" . $row['weaponCategory'] ."'>" . $row['weaponCategory'] ."</option>";
    } ?>
    </select>
    <input type="Submit" value="Change">
</form>

<?php
}else{
    echo 'No entry found. <a href="javascript:history.back()">Go back</a>';
}
?>

Right off the bat, you're mixing `mysqli_` with `mysql_` so that's a no go. You're connecting with `mysqli_` but querying with `mysql_` — Funk Forty Niner, Feb 14 '14 at 22:34
What's wrong with an array? Just change your `if` to `if ($row = mysql_fetch_array($query))` and remove the while loop. You have all the information you need in the `$row` variable. — jeroen, Feb 14 '14 at 22:39
(FYI)=> There are just too many things wrong with this code, which is why I didn't bother putting in an answer. — Funk Forty Niner, Feb 14 '14 at 22:40
Hey my bad i completely forgot I was connecting with mysqli. I know, retarded. Anyways I have taken all your advice but still no luck. Also, if I use simply just the `if` statement as mentioned by @jeroen the page is just blank. Please excuse my ignorance. In case any one had any other thoughts on this I have updated the code to show changes made. — doriansm, Feb 14 '14 at 22:54
I don't see any changes here. Edit: ah, here we go, it just kicked in. — Funk Forty Niner, Feb 14 '14 at 22:55
You'd still need to change to the mysqli variant of the function (I just copied your code to illustrate)... — jeroen, Feb 14 '14 at 22:55
Instead of `if(mysqli_num_rows($query)>=1){` can you try `if(mysqli_num_rows($query) >0){` @Aldentec or `if(mysqli_num_rows($query) ==1){` — Funk Forty Niner, Feb 14 '14 at 22:57
`if(mysqli_num_rows($query) >0){` did the trick. Thank you so much @fred -ii- — doriansm, Feb 14 '14 at 22:58
Plus, you may also need to add `$weaponName = $_POST['weaponName'];` @Aldentec — Funk Forty Niner, Feb 14 '14 at 23:00
Just did, glad it got resolved :) @Aldentec and you're welcome. — Funk Forty Niner, Feb 14 '14 at 23:01

echo_Me · Answer 1 · 2014-02-14T22:41:37.750

1

change this

 $query = mysql_query("SELECT * FROM weapons limit 1");

to

 $query = mysqli_query("SELECT * FROM weapons limit 1");

BUT omg all your code is mysql while you connected by mysqli !! .

edited Feb 14 '14 at 22:41

answered Feb 14 '14 at 22:36

echo_Me

37,078
5
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http://stackoverflow.com/questions/21790692/form-not-displaying-in-php#comment32970742_21790692 – Funk Forty Niner Feb 14 '14 at 22:36
Then for ` – Funk Forty Niner Feb 14 '14 at 22:39
Thanks, I put up some rather long ones sometime lol - and you seem like a nice guy too. I'm from Canada, just outside Ottawa, and you? and thanks for the badge :) – Funk Forty Niner Feb 14 '14 at 23:02
I deleted comments also. Sweden, nice place, from the many photos, documentaries I've seen. Somewhere I'd want to visit. – Funk Forty Niner Feb 14 '14 at 23:10
There are many beautiful places here. One of which I enjoy visiting (been there 2 times) is the Maritimes. Nova Scotia, Gaspesie, Prince Edward Island etc. and Quebec is a beautiful province. The people here are very nice :) – Funk Forty Niner Feb 14 '14 at 23:32
Yes that's what I meant, very beautiful surroundings. Mountains, rivers, lakes :) – Funk Forty Niner Feb 14 '14 at 23:33
Oh ok lol I meant that the people are friendly, yet yes... some are beautiful too hahaha a bit of both ;-) – Funk Forty Niner Feb 14 '14 at 23:38
You too, have a great day/evening. It's evening here now. Take care my friend :) – Funk Forty Niner Feb 15 '14 at 00:04

score 1 · Answer 2 · answered Feb 14 '14 at 22:37

1

You're mixing functions

mysqli_connect("localhost","username","password","db_name");

Won't work with

mysql_query("SELECT * FROM weapons limit 1");

Try

$query = mysqli_query($con, "SELECT * FROM weapons limit 1");

And then

if($query->num_rows >= 1)

answered Feb 14 '14 at 22:37

Machavity

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score 1 · Accepted Answer · answered Feb 14 '14 at 23:00

1

As requested by OP (from comment conversations)

Instead of

if(mysqli_num_rows($query)>=1){

use

if(mysqli_num_rows($query) >0){

answered Feb 14 '14 at 23:00

Funk Forty Niner

74,450
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score 0 · Answer 4 · answered Feb 14 '14 at 22:38

0

You connect with mysqli, which is fine. THEN, you attempt to run queries via mysql. Those are two separate extensions. You can't mix them as they won't "communicate" with one another. Stick to mysqli.

answered Feb 14 '14 at 22:38

Wayne Whitty

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Form not displaying in PHP

edit.php

4 Answers4