如何将mysqli查询结果值分配给php变量？

我希望代码回显某个值，即表中的列数。但我没有看到任何印刷品。我做错了什么？

您的查询是错误的。你可以这样做：-

<?php                
    $con=mysqli_connect("localhost","user","pass","db");

    // Check connection
    if (mysqli_connect_errno()){
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    // Perform queries 
    $result = mysqli_query($con,"SELECT COUNT(*)\n"
          . "FROM INFORMATION_SCHEMA.COLUMNS\n"
          . "WHERE table_name = \'CustomersTable\'");       

    $something = mysqli_fetch_assoc($result);

    echo $something;             

    mysqli_close($con);
?>

---

do
$result=mysqli\u查询（$con，“从信息\u SCHEMA.COLUMNS\n”“WHERE table\u name=\'CustomersTable\”）中选择COUNT（*）\n”“”）或die（mysqli\u错误（$con））
并检查是否有错误从语句中删除
\n
删除反斜杠并在
之前添加一个空格，其中
和
来自
。还可以使用
mysqli\u error
检查语句中的错误。使用
\'
可以逃避什么？
 <?php       

    $con=mysqli_connect("localhost","user","pass","db");

    if (mysqli_connect_errno()){
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $result = mysqli_query($con,"SELECT COUNT(*) as total_count FROM INFORMATION_SCHEMA.COLUMNS WHERE table_name = 'CustomersTable'") or die(mysqli_error($con));

    $something = mysqli_fetch_assoc($result);

    echo $something['total_count'];//or do var_dump($something);

    mysqli_close($con);
?>