Question

我有一个不起作用的多重查询。当我在前一页按下提交时，我进入一个空白页面，表中没有插入任何内容＆＃34; answer_det＆＃34;在我的数据库和文本＆＃34;信息存储成功＆＃34;没有出现。我做错了什么？

＆＃13;

<?php
ini_set('display_errors',1);
error_reporting(E_ALL);

if(mysqli_connect_errno())
{
	echo mysqli_connect_error();
}

$pid5 = $_POST['pid4'];
    
$aid1 = $_POST['a1'];
$aid2 = $_POST['a2'];
$aid3 = $_POST['a3'];
$aid4 = $_POST['a4'];    
    
$answ1 = $_POST['ans1'];
$answ2 = $_POST['ans2'];
$answ3 = $_POST['ans3'];
$answ4 = $_POST['ans4'];
    
$point1 = $_POST['pointset1'];  
$point2 = $_POST['pointset2'];  
$point3 = $_POST['pointset3'];  
$point4 = $_POST['pointset4'];

$que = "INSERT INTO answer_det VALUES('$pid5','','$aid1','$answ1','$point1')";
$que .= "INSERT INTO answer_det VALUES('$pid5','','$aid2','$answ2','$point2')";
$que .= "INSERT INTO answer_det VALUES('$pid5','','$aid3','$answ3','$point3')";
$que .= "INSERT INTO answer_det VALUES('$pid5','','$aid4','$answ4','$point4')";

$run = mysqli_multi_query($mysqli,$que);
if($run)
{
	echo "<br>Information stored successfully";

}
else
{
	echo mysql_error();
}


?>

＆＃13;

Answer 1

如果您只需要插入这些数据，则无需多次查询，使用此命令将代码作为一个查询运行：

$que = "INSERT INTO answer_det VALUES('$pid5','','$aid1','$answ1','$point1'), ";
$que .= "('$pid5','','$aid2','$answ2','$point2'), ";
$que .= "('$pid5','','$aid3','$answ3','$point3'), ";
$que .= "('$pid5','','$aid4','$answ4','$point4');";

$run = mysqli_query($mysqli, $que);

Answer 2

除了我的评论（添加分号）：

$que = "INSERT INTO answer_det VALUES('$pid5','','$aid1','$answ1','$point1');";
$que .= "INSERT INTO answer_det VALUES('$pid5','','$aid2','$answ2','$point2');";
$que .= "INSERT INTO answer_det VALUES('$pid5','','$aid3','$answ3','$point3');";
$que .= "INSERT INTO answer_det VALUES('$pid5','','$aid4','$answ4','$point4');";

$run = mysqli_multi_query($mysqli,$que);

请参阅清楚显示的PHP manual：

执行一个或多个由a连接的查询分号。

多个查询无法执行

2 个答案: