Php mysql的json输出不正确
我从MySQL发出了一个请求,下面的代码不起作用:Php mysql的json输出不正确,php,mysql,Php,Mysql,我从MySQL发出了一个请求,下面的代码不起作用: $email =$_POST['email']; $parola =$_POST['parola']; $rez = mysqli_query($con, "SELECT * FROM utilizatori WHERE email='$email' AND parola='$parola'"); if($rez){ $succes = 1; } else { $succes = 0; } $data = array("
$email =$_POST['email'];
$parola =$_POST['parola'];
$rez = mysqli_query($con, "SELECT * FROM utilizatori WHERE email='$email' AND parola='$parola'");
if($rez){
$succes = 1;
} else {
$succes = 0;
}
$data = array("succes" => $succes);
echo json_encode($data);
在《邮递员》中,它向我展示了这一点:您遗漏了数据变量,或者您的查询没有返回任何输出。请尝试在
Phpmyadmin
或CLI中使用MySQL查询。如果正常,则使用如下代码:
$email =$_POST['email'];
$parola =$_POST['parola'];
$rez = mysqli_query($con, "SELECT * FROM utilizatori WHERE email='$email' AND parola='$parola'");
while($result = mysqli_fetch_array($rez))
{
$data[] = $result;
}
if(mysqli_num_rows($rez) != 0) {
$data = array("success" => "Success", "data" => $data );
} else {
$data = array("success" => "Failed", "data" => array() );
}
echo json_encode($data);
这意味着您的查询失败,请回显
mysqli_error($con)
以查看错误消息。数据变量在哪里?执行rez=mysqli_查询($con,“从实用工具中选择*,其中email='$email'和parola='$parola')或死亡(mysqli_error($con))代码>并检查错误另外,您的SQL对SQL注入开放,请使用prepare语句以提高安全性请参阅: