Question

我有这个代码允许我在数据库中插入一些我选择的日期，现在这个只插入一个值，而不是全部，就像它停在第一个值一样。

你能告诉我我做错了什么吗？

谢谢！

$job_id=$_REQUEST['job_id'];

$dates = explode(",", $_POST['altField']);
foreach($dates as $date){
$sql="INSERT INTO date (job_id,date) VALUES('$job_id','$date')";
mysql_query($sql);
}

Answer 1

$job_id=$_REQUEST['job_id'];

$dates = explode(",", $_POST['altField']);
$values_arr = array();    
foreach ($dates as $date) {
    $values_arr[] = "('" . $job_id . "','" . $date . "')";
}
$values = implode(", ", $values_arr);

$sql="INSERT INTO date (job_id,date) VALUES $values";
mysql_query($sql);

Answer 2

$job_id=$_REQUEST['job_id'];

$dates = explode(",", $_POST['altField']);
foreach($dates as $date){
    $values .= '("'.$job_id.'","'.$date.'"),';
}
$sql="INSERT INTO date (job_id,date) VALUES " . substr($values, 0, -1);
mysql_query($sql);

插入可以有多个值集，因此您只需要在foreach中设置值，然后在其外部进行查询。你也可以这样做，得到相同的结果，

$job_id=$_REQUEST['job_id'];

$dates = explode(",", $_POST['altField']);
foreach($dates as $date){
    $values[] = '("'.$job_id.'","'.$date.'")';
}
$sql="INSERT INTO date (job_id,date) VALUES " . implode(',',$values);
mysql_query($sql);

<强>提醒

始终清理您的输入。不这样做可能导致SQL注入并导致重大问题。看看this还没有维护mysql，建议切换到MySqli或PDO。

Answer 3

尝试：

$sql = "INSERT INTO date SET job_id='".$job_id."', date='".$date."'";
mysql_query($sql);

一次在数据库中插入多个数据

3 个答案: