json_encode布尔返回错误的值

Question

我在写一个jQuery的登录。如果php登录脚本（checklogin.php）回应“true”，那么我的jquery页面应该会发出警报。当我运行一个login.php（不使用jquery）时，它将请求发送到checklogin.php，它反映了“true”，但是如果我使用我的jquery登录脚本将登录信息发送到checklogin.php，则警报I跟踪回说“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"; 
} 
?> 

所以长话短说，当我尝试从一个普通的PHP页面的价值我跟踪是真的，当我试图从一个jQuery页面的价值我迹是假的登录到登录;

编辑：

<?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 
     } 
     }); 
}); 

随着json_encode，布尔的值返回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'是一个字符串，你想一个布尔值：

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

此外，从未写== true。

if(data==true) {

---

好了，现在的问题是：

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

不要echo true，echo json_encode($json)。