0
我试图创建一个页面为一个ivent管理persone
,我得到2个错误SELECT ID第2行
1错误JOIN时间persone.id = time.p” :
每当我刷新页面时,我在我的数据库中找到一个空的数据。
第二个错误:
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 '(SELECT id from 'persone' LEFT JOIN time on persone.id=time.p' at line 2
感谢您的帮助
这是我的代码
<?php
include('config.php');
$nom = (!empty($_POST['nom']))?$_POST['nom']:"";
$prenom = (!empty($_POST['prenom']))?$_POST['prenom']:"";
$cin = (!empty($_POST['cin']))?$_POST['cin']:"";
$tel= (!empty($_POST['tel']))?$_POST['tel']:"";
$cat= (!empty($_POST['categore']))?$_POST['categore']:"";
$date= (!empty($_POST['d_donation']))?$_POST['d_donation']:"";
$time=(!empty($_POST['time']))?$_POST['time']:"";
$sql="INSERT INTO persone(nom, prenom, cin, tel, categore ,d_donation) value ('$nom','$prenom', '$cin','$tel','$cat','$date')
(SELECT id from 'persone'
LEFT JOIN time
on persone.id=time.p_id)";
if (mysqli_query($con, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($con);
}
exit;
?>
此代码的用途是什么?(来自'persone'的SELECT编号为 'persone.id = time.p_id'的LEFT JOIN时间 '')。如果您试图插入记录,这不是必需的。 – gvmani
你想插入数据或更新persone.id的数据? – scaisEdge
引用是字符串,反引号是表/列。您可以接受SQL注入。把所有的值放到'select'中,或者把'select'作为子查询放在'values'中....虽然它看起来像是你匹配的列和值的数量。你是否试图插入该人的ID在某处或选择后? – chris85