Question

此代码将从另一个PHP代码接收id和区域，并且我确信它已正确接收它们。此代码用于更新id等于用户输入的id的区域。我运行代码，我从这个语句中得到了SQL语法错误。

$sql = "UPDATE complain SET area = ".$area."WHERE num = ".$id."";

任何人都知道出了什么问题？

<?php

$area = $_GET['area']; 
$id = $_GET['id']; 

echo $area;
echo $id;

//$Area = $_POST['variable'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sfa";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

 $sql = "UPDATE complain SET area = ".$area."WHERE num = ".$id."";



if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

?>

Answer 1

在Where子句之前添加一些空格，使您的查询像

UPDATE complain SET area = ".$area." WHERE num = ".$id."

Answer 2

".$area."WHERE

之间缺少空格

结果是

UPDATE complain SET area = 2WHERE num = 0815

Answer 3

尝试使用此查询。它工作正常并更新记录 -

＆＃34;更新投诉SET区=＆＃39;。$ area。＆＃39; WHERE num =＆＃39;。$ id。＆＃39;＆＃34;

Answer 4

以下查询应该有效：

$scope.addCart = function() {
    $http({
        method  : 'POST',
        url     : 'server/menu/add_cart',
        headers : { 'Content-Type' : 'application/json' },
        data    : JSON.stringify({ id: $scope.id ,menu: $scope.menu , harga:$scope.harga })
    }).success(function(data) {
        console.log(data);
    });
}

希望它有所帮助！

UPDATE中的SQL语法错误

4 个答案: