Question

我正在尝试插入这样的数据

function submitData() {
//Posting to ContactDB with JSON format
$.post("contactDB.php",
    JSON.stringify({ 
          name: $("#name").val(),
          email: $("#email").val(),
          phone: $("#phone").val(),
          message: $("#message").val() 
    }),
    function(usrava){
        // if data is inserted successfully than show insert success message in Result div
        if(usrava=='Data Inserted')
        {
            $("#result").fadeTo(200,0.1,function(){ 
                $(this).html('Your message is successfully saved with us. We will get back to you soon.').fadeTo(900,1);
            });     
        }
        //else show the error 
        else
        {
            $("#result").fadeTo(200,0.1,function(){ 
                $(this).html(usrava).fadeTo(900,1);
            });
        }
    });
}

的 ContactDB.php

<?php
    $mysqli = new mysqli("localhost", "root", "", "contactDB"); //Connection to the Database
    //If Error than die
    if (mysqli_connect_errno()) { 
        die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error());
        //Echo for the response 
        echo "Data base connection NOT Successful. Please get the assistance from your Administrator.";
        //Should not go out if not connected
        exit();
    }
    //data received in json. decoded it to an array
    $data = json_decode(file_get_contents('php://input'), true);
    //Create a insert Command using implode as the data is already in the array
    $insert = "INSERT INTO contact(Name,Email,Phone,Message) VALUES ('" .implode(",",$data)."')";
    $mysqli->query($insert);
    //Close the connection
    $mysqli->close();
    // Echo for the response
    echo "Data Inserted";
?>

问题是我$mysqli->query($insert); 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 Rajesh, at line 1发现错误<form method="post"> <input type="text" name="Name" id="Name"> <input type="submit" value="Submit"> </form>

我该怎么做让它运行。我已经尝试了一切，但找不到任何有效的解决方案。请帮忙！！提前致谢

Answer 1

有一些env env env值。您需要正确使用make -p。

string

<强>解释

您正在使用的代码生成类似于 - '的语法错误。它们都是VALUES ('" .implode("','",$data)."')"，需要包含在VALUES ('name, email, phone, message')中。

string - 将正确包装它们。它会生成 - '

Answer 2

函数implode将使用字符串连接数组元素。现在你的函数是implode(",",$data)，这将使字符串像这样

NameData,EmailData,PhoneData,MessageData

然而，你会想要像这样做

'NameData','EmailData','PhoneData','MessageData'

要做到这一点，你需要使你的内爆函数看起来像这个implode("','",$data)，并且第一个和最后一个'已经从你编写的代码中进入。

Answer 3

试试这个：

$postdata =  implode("','", $data) ;

$insert = "INSERT INTO contact(Name,Email,Phone,Message) VALUES ('$postdata')";

使用JSON插入时出现问题

的 ContactDB.php

3 个答案: