php - mysqli_query() 期望参数 2 为字符串

Question

我遇到了错误

Warning: mysqli_query() expects parameter 2 to be string, object given in C:\Program Files (x86)\EasyPHP-DevServer-13.1VC9\data\localweb\login.php on line 10Error:

在执行查询时

<?php
$con=mysqli_connect("127.0.0.1","root","","forms");
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
  
  $sql=mysqli_query($con,"SELECT * FROM register Where username='$_POST[username]' AND password='$_POST[password]'");
  
  if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }
echo "login success";

mysqli_close($con);
?>

Answer 1

if(!mysqli_query($con,$sql)) 不好，因为 $sql 是对象，不是字符串

$sql 是一个对象，如 mysql_query

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

所以你不必做另一个mysqli_query，只需按以下方式修改你的if

if(!$sql)
{
  //do something
}

如果您需要检查查询失败，这将起作用，否则即使您的查询返回了 0 行，它也可能返回 mysqli_result，所以请注意。