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I am using PHP and MYSQL to make a search field and it is working fine. But when I type % and click search, I am getting all rows of the column description from the database. Here is the code:
  <!-- Main Content -->
  <form action="test.php" method="GET">
  <div class="container mb-5">
    <div class="row">
      <div class="col-lg-6">
        <label for="search" class="mb-2 mt-4 fw-bold d-flex justify-content-center" style="font-size: 20px; color: #6f42c1;">Search</label>
        <div class="input-group mb-4">
        <input type="search" class="form-control" name="search" value="<?=(isset($_GET['search'])?$_GET['search']:'');?>">
        <button type="submit" class="btn btn-outline-secondary"></button>
      </div>
      <div class="d-flex justify-content-center mb-4">
      <canvas class="bg-light" id="myCanvas" style="border:1px solid #dee2e6;"></canvas>
    </div>

      <?php
      if(isset($_GET['search']) && !empty(trim($_GET['search']))){
          $condition = '';
          $param = explode(" ", sanitize($_GET['search']));
          foreach($param as $text){
            $condition .= "description LIKE '%" . mysqli_real_escape_string($db, $text) . "%' AND ";
          }
          $condition = substr($condition, 0, -4);
          $sql = "SELECT * FROM customers WHERE " . $condition;
          $result = $db->query($sql);
          if(mysqli_num_rows($result) > 0){
            while($row = mysqli_fetch_assoc($result)){ ?>

              <div>
              Description: <?=html_entity_decode(str_replace("p", "span", $row['hadeeth']));?>
              <hr>
            </div>

            <?php }
          }else{
            echo 'No results';
          }
      }
    ?>
</div>
</div>


What I have tried:

If I search normally, I am getting the correct results. Also, if search field is empty and I click on search I am not getting any results, which is also correct. But if I type % and then I click on search button, I am getting all rows from the column description which is not what I need. Why is that happening and what's the solution?
Posted
Updated 18-Mar-23 20:18pm
v3

The '%' character in an SQL LIKE clause is a wildcard: "match any number of any characters including none" - it's like "*" in a file name under Windows: "*.*" matches every file name with every extension.

If you want to allow the user to search for a literal percent character, it needs to be escaped: MySQL 8.0 Reference Manual: 9.1.1 String Literals[^]
But ... do remember that your code almost certainly adds '%' to the front and end of the search string the user enters! Escape those and all searches will return no results.
 
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Comments
FRS4002 19-Mar-23 14:40pm    
Ok thanks for explaining the issue. I think you are right ... But I didn't understand the solution ... I checked the link you sent, there is a table 9.1, that it says I should use \%, I don't know if I understand or no. So, I changed the code to this:
$condition .= "description LIKE '\%" . mysqli_real_escape_string($db, $text) . "\%' AND ";
But the search stopped working ... I am not getting any results even if I type an existing word ... Could you please help me what to do ??
OriginalGriff 19-Mar-23 14:45pm    
If you escape *all* the '%' characters, there are no wild cards left...
You need to leave the ones you add alone and escape all the ones in the data the user enters only.
FRS4002 19-Mar-23 14:55pm    
Ok, what I understand from you is to leave % on both sides and add \%. I tried that here:
$condition .= "description LIKE '%\%" . mysqli_real_escape_string($db, $text) . "%' AND ";
It didn't work too ... Sorry, I am facing difficulties on how to apply the escape on the code ... Is it even doing \% right ?? Could you please help me apply the escape on the code ??

OriginalGriff 19-Mar-23 15:29pm    
Oh, come on ... read what I typed ...

If the user entered "100% dairy free" then the search string you want to end up with is "%100\% dairy free%" so that it matches "Soya Milk 100% dairy free 500ml", "100% dairy free soya milk 500ml", and "500ml Soya Milk 100% dairy free" - but not "Soya Milk 100 percent dairy free 500ml"
So you need to modify what the user typed to escape the '%' character ...
FRS4002 19-Mar-23 21:28pm    
Done!! I just added:
$param = str_replace('%', '\\%', $param);
above foreach loop. Thanks alot!
Have a look at: SQL Wildcard Characters[^]
 
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