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Here is my code:

The error says it is on line 13. Line 13 is where this code is at:
if (mysqli_num_rows($stock_query)==0) {


    //if categoryID is not set, redirect back to index page
    if(!isset($_GET['tadilatID'])) {
// Select all hizmetveren items belonging to the selected categoryID
  $stock_sql="SELECT usta.ustaID, usta.namelastname,, usta.phonenumber,, AS tadname FROM usta JOIN tadilat ON usta.tadilatID=tadilat.tadilatID WHERE usta.tadilatID=".$_GET['tadilatID'];
  if($stock_query=mysqli_query($dbconnect,$stock_sql)) {
  ---->>>> Line 13: Error Line -> if (mysqli_num_rows($stock_query)==0) {
    echo "Pek Yakinda Hizmetinizde";
  } else {

      <h1><?php echo $stock_rs['tadname']; ?></h1>
    <?php do {
line-height: 0.2;
width: 300px;

        <div class="item">
          <a href="index.php?page=item&ustaID=<?php echo $stock_rs['ustaID']; ?>"><p><?php echo $stock_rs['namelastname']; ?></p>
          <p>E-Mail:🌐<?php echo $stock_rs['email']; ?></p>
          <p>Telefon: 📞<?php echo $stock_rs['phonenumber']; ?></p>
          <p>Website: 🌐<?php echo $stock_rs['website']; ?></p></a>



What I have tried:

I tried changing line 13 to ==1 instead of ==0. Or tried to switch else code with if else code. For example if previously it was: if x, else y, I switched it to if y, else x. Didn't work.

I have no idea at this point. Please help! Thank you very much.
Updated 10-Apr-21 21:29pm
Richard Deeming 12-Apr-21 6:45am    
Your code is vulnerable to SQL Injection[^]. NEVER use string concatenation to build a SQL query. ALWAYS use a parameterized query.
PHP: SQL Injection - Manual[^]

1 solution

Share this answer
zebestd 11-Apr-21 3:45am    
I don't understand. Could you please elaborate?
Richard MacCutchan 11-Apr-21 4:05am    
Look at the documentation:
Return Values ¶For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or false on error.

So a return value of false is telling you that your SELECT clause produced an error.

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