如何检查数据从php页面返回到ajax成功函数

我想检查成功函数数据使用if循环在ajax下面是任何一个帮助我的代码。如何检查数据从php页面返回到ajax成功函数

如果echo部分返回成功警报将为我带来这个问题。

下面是javascript部分和php部分

$(document).ready(function() { 
      $("#newuserBtn").on("click", function() { 
       $("#regform").show(); 
       $("#loginform").hide(); 
      }); 

      $("#submitBtn").on("click", function() { 
       var uid = $("#uid").val(); 
       var pin = $("#inputPassword").val(); 
       $.ajax({ 
        type: "POST", 
        url: "validation.php", 
        data: { 
         uname: uid, 
         pwd: pin 
        }, 
        success: function(value) { 
         if (value == "success") { //problem here 
          alert(); 
         } 
        } 
       }); 

      }); 

// php code 


<?php 
$servername = "localhost"; 
$username = "root"; 
$password = "ipad7*"; 
$db_name = "users"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $db_name); 

// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
//Gets username value from the URL 
$uname = $_POST['uname']; 
$pwd = $_POST['pwd']; 

//Checks if the username is available or not 
$query = "SELECT pin FROM users_details WHERE uid = '$uname'"; 
$result = mysqli_query($conn, $query); 

if (mysqli_num_rows($result) > 0) { 
    // output data of each row 
    while ($row = mysqli_fetch_assoc($result)) { 
     if ($row["pin"] == $pwd) { 
      echo "success"; 
     } else { 
      echo "login fail"; 
     } 
     //echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>"; 
    } 
} else { 
    //echo "login fail"; 
} 
$conn->close(); 

//echo "fff"; 
//echo "$result"; 


?>

来源

2015-06-11 MohanaRajesh

尝试使用'的console.log（）'来打印你的'ajax'响应，并检查你做了什么在响应 –

我试过它在控制台显示成功，但当我检查使用如果（价值==“成功”）{/ /这里的问题 alert（）; } alert noting – MohanaRajesh

我试过布尔值insted字符串，现在工作“if（value == true）{}”但我想字符串值分隔的解决方案也 – MohanaRajesh

你检查是否有在PHP脚本收盘?>标签后没有多余的线端？如果存在，则在console.log中显示该字符串仍然可以看起来不错，但不会相同。
这是一个好主意，完全删除结束标记?>，以避免这样的错误。

来源

2015-06-11 10:40:31

试试这个：在json编码发送您的responce（在游码值）格式

<script> 
    $(document).ready(function() { 
      $("#newuserBtn").on("click", function() { 
       $("#regform").show(); 
       $("#loginform").hide(); 
      }); 

      $("#submitBtn").on("click", function() { 
       var uid = $("#uid").val(); 
       var pin = $("#inputPassword").val(); 
       $.ajax({ 
        type: "POST", 
        url: "validation.php", 
        contentType: "application/json; charset=utf-8" 
        data: { 
         uname: uid, 
         pwd: pin 
        }, 
        success: function(value) { 
         console.log(value); 
         if (value == "success") { //problem here 
          alert(); 
         } 
        } 
       }); 

      }); 


</script> 

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = "ipad7*"; 
$db_name = "users"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $db_name); 

// Check connection 
if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 
//Gets username value from the URL 
$uname = $_POST['uname']; 
$pwd = $_POST['pwd']; 

//Checks if the username is available or not 
$query = "SELECT pin FROM users_details WHERE uid = '$uname'"; 
$result = mysqli_query($conn, $query); 

if (mysqli_num_rows($result) > 0) { 
    // output data of each row 
    while ($row = mysqli_fetch_assoc($result)) { 
     if ($row["pin"] == $pwd) { 
      // echo "success"; 
      echo json_encode("success"); 
      exit(); 
     } else { 
      // echo "login fail"; 
      echo json_encode("login fail"); 
      exit(); 
     } 
     //echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>"; 
    } 
} else { 
    //echo "login fail"; 
} 
$conn->close(); 

//echo "fff"; 
//echo "$result"; 


?>

来源

2015-06-11 10:43:40

如何检查数据从php页面返回到ajax成功函数

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