Question

问题：我可以将值传递给ajax方法并成功实现Select SQL（在网络中看到的结果），但它永远不会返回任何结果（我的结果是警报（＆＃34;成功）。问题是什么它？

HTML：

function selectupdate_Server(id)
    {
        var ID= id;
        console.log(ID);   --->Successfully logged 
        $.ajax({
            url: "abc.php",
            type: 'POST',
            data: {"ID": ID},
            dataType: 'json',
            success: function (data) {
                alert ('Success');   ------>Failed to alert

            }
        });

    }

PHP：

   $serverName = "localhost";
  $username = "root";
   $password = "root";
  $dbName = "noticeboard";
  $tbName1 = "notice";
   $conn = new mysqli($serverName, $username, $password, $dbName);

  checkStaffLogin();

function checkStaffLogin()
{
global $conn, $tbName1;
if ($conn->connect_errno)
{
    echo "Fail to connect MYSQL database : " . $conn->connect_errno;
}
else {
     $ID= $_POST["ID"];

    $sql = "SELECT *
                     FROM $tbName1 where ID= '$ID'";

    $result = $conn->query($sql);
   if($result)
 { 
    while($row =mysqli_fetch_assoc($result) )
    {
        $table_data[]=           array("Name"=>$row['Name'],"Password"=>$row['Password']);
    }

        echo json_encode($table_data);
}

Answer 1

尝试此查询

 $table_data=[];
 $sql = "SELECT * FROM ".$tbName1." where ID = ".$ID;

注意：使用preperd语句并将此标题附加到您的php header('Content-Type: application/json');

循环结果使用

success: function (data) {
               $.each(data,function(i,v){
                     console.log(v.Name)
               });
            }

注意数据应如下所示：

data: {ID: ID},

Answer 2

   <?php 
    $ID= $_POST["ID"];

        $sql = "SELECT * FROM ".$tbName1."where ID=".$ID;

        $result = $conn->query($sql);
        while($row =mysqli_fetch_assoc($result) )
        {
            $table_data[]=           array("Name"=>$row['Name'],"Password"=>$row['Password']);
        }

            echo json_encode($table_data);
?>

试试这个

Answer 3

在JS中，通过删除那些＆＃34; ID＆＃34;来检查数据的语法。请尝试以下

function selectupdate_Server(id)
    {
        var ID= id;
         $.ajax({
            url: "abc.php",
            type: 'POST',
            data: {ID: ID},
            dataType: 'json',
            success: function (data) {
                alert ('Success');  

            }
        });

    }

Answer 4

你试试这个Javascript函数

function selectupdate_Server(id)
{
    var ID= id;
    console.log(ID);   --->Successfully logged 
    $.ajax({
        type: 'POST',
        data: {ID: ID},
         url: "abc.php",
        dataType: 'json',
        success: function(resultss){
            try{
             alert ('Success'); 
            }catch(e)
            {
                alert(JSON.stringify(e)+"Catch error");//any Error in Your php file display alert.where,what error
            }

        },
        error: function(e){

         alert(JSON.stringify(e)+"error");//any Error in Your php file display alert.where,what error

        }
        });

}

这个脚本，你的php文件中的错误提醒你行no和你检查和更正的文件名

和你这样的Php代码

 checkStaffLogin();
 function checkStaffLogin()
 {
  $serverName = "localhost";
  $username = "root";
  $password = "root";
  $dbName = "noticeboard";
  $tbName1 = "notice";
  $conn = mysqli_connect($serverName, $username, $password, $dbName);

  if (mysqli_connect_errno($conn))
  {
   echo "Failed to connect to MySQL: " . mysqli_connect_error();
 }
else {
  $ID= $_POST["ID"];
  $table_data=array();
  $sql = "SELECT * FROM $tbName1 where ID= '$ID'";

  $result = mysqli_query($conn,$sql) or die(mysql_error());
  if($result)
  { 
  while($row =mysqli_fetch_assoc($result) )
  {
    $table_data[]= array("Name"=>$row['Name'],"Password"=>$row['Password']);
  }

    echo json_encode($table_data);
  }
 }
 }

你试试这个

ajax回调方法但返回失败

4 个答案: