Question

我一直收到这个错误：

Warning: mysqli_error() expects exactly 1 parameter, 0 given in /Applications/MAMP/htdocs/dbViewer.php on line 71

无法获取数据：

来自此代码

$sql = "SELECT DISTINCT TABLE_NAME FROM INFORMATION_SCHEMA.COLUMNS WHERE COLUMN_NAME IN ('Version') AND TABLE_SCHEMA = 'wp_plugin_db'";

 $result = mysqli_query($con,$sql);

 if(! $result )
{
  die('Could not get data: ' . mysqli_error());
}

 $arrayCount = 0;


 while ($row=mysqli_fetch_array($result)) 
 {

  $tableNames[$arrayCount] = $row[0];
  $arrayCount++;
 }

 foreach ($tableNames as $siteName) {
    $siteName =  mysqli_real_escape_string($con,$siteName);
    $sql="SELECT Plugin_Name, Version, WPVersion FROM `".$siteName."` ORDER BY   Plugin_Name";

    $result=mysqli_query($con,$sql);
    if(! $result )
    {
      die('Could not get data: ' . mysqli_error());
    }
    echo "Website Name:  $siteName ----  " ;
    while($row=mysqli_fetch_array($result,MYSQLI_ASSOC))
    {
        echo "  Plugin Name :{$row['Plugin_Name']}   ".
             "  Version : {$row['Version']}  ". 
             "  Wordpress Version : {$row['WPVersion']}  ".
             "   |    ";


    } 
    echo "<br>";
 }

mysqli_close($con);

错误显示line 71 die('Could not get data: ' . mysqli_error());

真正的错误发生在$sql="SELECT Plugin_Name, WPVersion, Version FROM ".$siteName." ORDER BY Plugin_Name";

我知道这一点，因为当我从selec中取出WPVersion并注释掉该元素的echo时，它可以正常工作。请帮助我似乎无法解决这个问题，是语法吗？我觉得很傻哈哈。

提前谢谢！

Answer 1

你真的读错了吗？你忘了将链接参数传递给mysql，ep>

mysqli_error($con);
             ^^^^--- missing

尝试吐出数据库元素时出现Mysql错误

1 个答案: