php - MySQL 语法错误显示 "near ' 1'"但我的查询中没有 '1'

Question

我第一次使用 PHP 和 MySQL 数据库创建一个网站，并收到一个我不明白的语法错误。我只想删除 tblreparation 中的所有内容具有一定的ID以及 rblrepstat 中的所有内容与此相同ID 。我正在使用这段代码:

<?php
    $con=mysqli_connect("localhost","root","MYPASS","repair");
    $ID = $_REQUEST['ID'];
    $sql = mysqli_query($con, "DELETE FROM tblreparation WHERE ID = {$ID}");
    if (!mysqli_query($con,$sql)) {
      die('Error: ' . mysqli_error($con));
    }
    $sql = mysqli_query($con, "DELETE FROM tblrepstat WHERE repID = {$ID}");
    if (!mysqli_query($con,$sql)) {
      die('Error2: ' . mysqli_error($con));
    }
    echo "1 record deleted";
    mysqli_close($con); 
?>

这是我收到的错误:

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1' at line 1

接近 1？我什至没有看到“1”...

Answer 1

您为每个查询调用 mysqli_query 两次。

第二次调用它时，您实际上将资源作为查询参数传递，这会导致您收到错误。

尝试将代码更改为:

<?php
    $con=mysqli_connect("localhost","root","MYPASS","repair");
    $ID = $_REQUEST['ID'];
    $sql = "DELETE FROM tblreparation WHERE ID = {$ID}";
    if (!mysqli_query($con,$sql)) {
      die('Error: ' . mysqli_error($con));
    }
    $sql = "DELETE FROM tblrepstat WHERE repID = {$ID}";
    if (!mysqli_query($con,$sql)) {
      die('Error2: ' . mysqli_error($con));
    }
    echo "1 record deleted";
    mysqli_close($con); 
?>