Question

我需要帮助才能将数据插入mysql。我无法在下面的代码中找到为什么数据没有插入数据库的问题。当我提交表单时，会显示以下消息：Notice: Undefined variable: asd in...

<?php
$q = mysql_query("select `cus_id`, `date`, `mobile`, 
    least(`t1`, `t2`, `t3`, `t4`) as min 
    from `table1`");

while($r = mysql_fetch_assoc($q)){

$asd = array(`cus_id` =>$r['cus_id'], 
             `mobile` =>$r['mobile'], 
             `date`   =>$r['date'], 
             `credit` =>$r['min']);
}

$sql = array();

foreach((array)$asd as $row){

$sql[] = '("'.mysql_real_escape_string($row['cus_id']).'", 
                                     '.$row['mobile'].', 
                                     '.$row['date'].', 
                                     '.$row['min_total'].'
           )';
}

$stmt = mysql_query('INSERT INTO `single` (`cus_id`, `mobile`, `date`, `credit`) 
                     VALUES '.implode(',', $sql));                                    

if(!$stmt){ 
echo "error". mysql_error();
}else{
$_SESSION['s']="Payment successfully saved";
header('location:final.php');
}
?>

感谢您的帮助。

Answer 1

如果我将上述查询转换为mysqli，则代码如下：

<?php

$link = mysqli_connect('localhost', 'root', '', 'mumbai');
$q = "select `cus_id`, `date`, `mobile`, 
      least(`t1`, `t2`, `t3`, `t4`) as min 
      from `table1`";

$asd = array(); 
$result = mysqli_query($link, $q);
while($r = mysqli_fetch_assoc($result)){

$asd = array(`cus_id` =>$r['cus_id'], 
             `mobile` =>$r['mobile'], 
             `date`   =>$r['date'], 
             `credit` =>$r['min']);
}

$sql = array();

foreach((array)$asd as $row){

$sql[] = '("'.mysqli_real_escape_string($row['cus_id']).'", 
                                 '.$row['mobile'].', 
                                 '.$row['date'].', 
                                 '.$row['min_total'].'
       )';
}
$stmt = mysqli_prepare($link, "INSERT INTO single 
               (`cus_id`, `mobile`, `date`, `time`, `credit`) VALUES (?,?,?,?,?)");

mysqli_stmt_bind_param($stmt, 'isssi', implode(',', $sql));                  

mysqli_stmt_execute($stmt);

if(!$stmt){ 
echo "error". mysqli_error($link);
}else{
$_SESSION['s']="Payment successfully saved";
header('location:final.php');
}
?>

Answer 2

在while循环之前添加$asd变量。

$asd = array();

while($r = mysql_fetch_assoc($q)){
....

然后在您的foreach中，您可以删除(array)

foreach($asd as $row){
....

不再Notice: Undefined variable: asd in...

从数组中插入多行

2 个答案: