Question

我有问题让这个工作。我想从数据库中选择行并将其呈现为列表。这是我的代码：

<ul class = "list">
  <li>
  <?php
    $x = 'some_string';
    $title = 'title';

    $dbc = mysqli_connect('localhost', 'root', '', 'database') or die('Error connecting');
    $query = "SELECT * FROM table WHERE column ==$x ORDER BY procenat DESC LIMIT 5";
    $data = mysqli_query($dbc, $query);
  ?>
   <div id = "listdiv">
  <?php
    while ($row = mysqli_fetch_array($data)) {
  ?>
    <h4><?php echo $row[$title]; ?> </h4>
  <?php 
    }
    mysqli_close($dbc);
  ?>
   </div> </li> </ul>

我尝试使用LIKE'％$ x％'并且效果不佳。我试过...... column = $ x

Answer 1

您需要使用单个等号和引号。

    $query = "SELECT * FROM table WHERE column = '$x' ORDER BY procenat DESC LIMIT 5";

Answer 2

您应该像这样使用LIKE运算符

$query = "SELECT * FROM table WHERE column_name LIKE '%$x%' ORDER BY procenat DESC LIMIT 5

Answer 3

更改查询行，如下所示

$query = "SELECT * FROM table WHERE column = '".$x."' ORDER BY procenat DESC LIMIT 5";

Answer 4

您的查询中有两个错误只是尝试这个。

将$x替换为'$x'
将==替换为=
运行您的查询

$x = 'some_string';

$query = "SELECT * FROM table WHERE column ='$x' ORDER BY procenat DESC LIMIT 5";

Mysql查询无法正常工作：SELECT * FROM＆＃39; table＆＃39;在哪里＆＃39;列＆＃39; =＆＃39; some_string＆＃39;

4 个答案: