Question

我试图编写的函数是用户在框中输入一个ID整数，JS读取该框并使用JSON和Ajax将该数据发送到PHP文件。此文件将接收ID并运行MySQL查询并将该答案返回到变量中以发送回JavaScript并放入表中。我不知道如何让PHP变量发送回JavaScript或它返回的格式，任何帮助将不胜感激。

$("#sid").click(function() {
    var selectID = document.getElementById("selectedID").value;//selects the ID it wants to
    var jsonSID = {"ID":selectID};
    var jsonSelectID = JSON.stringify(jsonSID);


    if(selectID=='')
        alert("Please select an ID to find");
    else{
        alert("goes to ajax")
        $.ajax ({
            type:"POST",
            url:"viewData.php",
            data: {selectData: jsonSelectID},
            dataType: 'json',
            success: function(data){

                console.log(data.id);
                console.log(data.name);
            },
            error: function(e){
                alert("Didn't work, refresh and it should work");
            }

        });
    }

});

PHP文件

         <?php
    $link = mysqli_connect("localhost", "root", "", "assignment5");
    if($link === false){
        die("ERROR: Could not connect. " . mysqli_connect_error());
    }
            $json = $_POST['selectData']; //Takes in the selected ID array
            $data = json_decode($json); //decodes the selected ID JSON array
            $selectedID = $data->ID; // this is the user selected ID

            $query = "SELECT * FROM datadump WHERE id = '$selectedID'"; //You don't need a ; like you do in SQL
            $result = mysql_query($query);
                $array = mysql_fetch_row($result);
                echo json_encode($array);
   ?>

Answer 1

注意：不推荐使用mysql id，切换到mysqli或PDO

如果您只想访问一个或两个变量变量，您可以在数组中查找并可以像这样执行：。

success: function(data id){
  alert(data);
  alert(id);
},

和如果你想要检索多条记录，你需要这样的东西：获取你的php有一个错误应该像这样：

  $query = "SELECT * FROM datadump WHERE id = '$selectedID'";
  $result = mysql_query($query);

  while($res = =mysql_fetch_array($result)){
    $data[] = $res; 
  }
$res['data'] = $data;
echo json_encode($res);

之后发送json对象，在前端你需要与$.each function

分开

使用AJAX将带有mySQL数据的PHP变量发送回javascript

1 个答案: