json_encode布爾返回錯誤的值

Question

我在寫一個jQuery的登錄。如果php登錄腳本（checklogin.php）迴應「true」，那麼我的jquery頁面應該會發出警報。當我運行一個login.php（不使用jquery）時，它將請求發送到checklogin.php，它反映了「true」，但是如果我使用我的jquery登錄腳本將登錄信息發送到checklogin.php，則警報I跟蹤回說「假」。

$('#login').click(function(e){ 
     $.getJSON("http://www.hedonsoft.com/game/login.php",{username: $("#username").val(), password: $("#password").val()},function(data){ 
     if(data==true){ 
      alert(data.boolean); 
      //$('#logindiv').slideUp(); 
      //$('#gamediv').show(); 
      //$('#registerdiv').hide(); 
      //$('#gameheader').html(data['username']); 
     }else{ 
      alert("false"); 
     } 
     }); 
}); 

<?php 
header('Access-Control-Allow-Origin: *'); 
$host="localhost"; // Host name 
$username=""; // Mysql username 
$password=""; // Mysql password 
$db_name="hedonsof_conflict"; // Database name 
$tbl_name="members"; // Table name 

// Connect to server and select databse. 
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 
mysql_select_db("$db_name")or die("cannot select DB"); 

// username and password sent from form 
$myusername=$_POST['myusername']; 
$mypassword=$_POST['mypassword']; 

// To protect MySQL injection (more detail about MySQL injection) 
$myusername = stripslashes($myusername); 
$mypassword = stripslashes($mypassword); 
$myusername = mysql_real_escape_string($myusername); 
$mypassword = mysql_real_escape_string($mypassword); 

$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; 
$result=mysql_query($sql); 

// Mysql_num_row is counting table row 
$count=mysql_num_rows($result); 

// If result matched $myusername and $mypassword, table row must be 1 row 
if($count==1){ 

// Register $myusername, $mypassword and redirect to file "login_success.php" 
session_register("myusername"); 
session_register("mypassword"); 
$json = array('boolean' => 'true', 'jsonUser' => $myusername); 
echo "true"; 
} 
else { 
echo "Wrong Username or Password"; 
} 
?> 

所以長話短說，當我嘗試從一個普通的PHP頁面的價值我跟蹤是真的，當我試圖從一個jQuery頁面的價值我跡是假的登錄到登錄;

編輯：

<?php 
... 
$sql="SELECT * FROM $tbl_name WHERE username='$myusername' and password='$mypassword'"; 
$result=mysql_query($sql); 

// Mysql_num_row is counting table row 
$count=mysql_num_rows($result); 

// If result matched $myusername and $mypassword, table row must be 1 row 
if($count==1){ 

// Register $myusername, $mypassword and redirect to file "login_success.php" 
session_register("myusername"); 
session_register("mypassword"); 
$json = array('boolean' => true, 'jsonUser' => $myusername); 
echo true; 
} 
else { 
echo "Wrong Username or Password"; 
} 
?> 

而我的jQuery ...

$('#login').click(function(e){ 
     $.getJSON("http://www.hedonsoft.com/game/login.php",{username: $("#username").val(), password: $("#password").val()},function(data){ 
     if(data){ 
      alert("true"); 
      //$('#logindiv').slideUp(); 
      //$('#gamediv').show(); 
      //$('#registerdiv').hide(); 
      //$('#gameheader').html(data['username']); 
     }else{ 
      alert(data); // shows null 
     } 
     }); 
}); 

隨着json_encode，布爾的值返回null

$('#login').click(function(e){ 
     $.getJSON("http://www.hedonsoft.com/game/login.php",{username: $("#username").val(), password: $("#password").val()},function(data){ 
      if(data.boolean == "true"){ 
      alert("true"); 
      //$('#logindiv').slideUp(); 
      //$('#gamediv').show(); 
      //$('#registerdiv').hide(); 
      //$('#gameheader').html(data['username']); 
      }else{   alert(data); 
      } 
     }); }); 

<?php 
... 
// Register $myusername, $mypassword and redirect to file "login_success.php" 
session_register("myusername"); 
session_register("mypassword"); 
$json = array('boolean' => true, 'jsonUser' => $myusername); 
echo json_encode($json); 
} 
... 
?> 

Answer 1

使用true，而不是'true'。 'true'是一個字符串，你想一個布爾值：

$json = array('boolean' => true, 'jsonUser' => $myusername); 

此外，從未寫== true。

if(data==true) {

---

好了，現在的問題是：

$json = array('boolean' => true, 'jsonUser' => $myusername); 
echo true; 

不要echo true，echo json_encode($json)。