检查返回的文本是真的还是假

Question

我有一个Ajax/PHP代码，我想检查返回的文本是真还是假：

$(".gets").click(function() { 

    // some details 

    $.ajax({ 
     type: "POST", 
     url: "step2.php", 
     data: dataString, 
     success: function(data) { 

      // This is the button to hide 
      $('.nums').fadeOut('fast'); 

     } // end function 
    }); // end ajax 

    return false; 
}); 

step2.php

if($num < 3) { 
echo 'please choose at least 3 nums<br />'; 
$hideButton = false; // hide show nums button 
} else {   
$hideButton = true; 
} 

在此先感谢

Answer 1

您必须回显$hideButton。这些信息将可再data

if($num < 3) { 
echo 'please choose at least 3 nums<br />'; 
    $hideButton = false; // hide show nums button 
} else {   
    $hideButton = true; 
} 
echo $hideButton; 

然后 - 改变你的脚本像下面

$.ajax({ 
    type: "POST", 
    url: "step2.php", 
    data: dataString, 
    success: function(data) { // data now has the value of $hideButton 
     if(data == 'true') { 
      // This is the button to hide 
      $('.nums').fadeOut('fast'); 
     } 
    } // end function 
}); // end ajax   

Answer 2

$(".gets").click(function() { 

// some details 

$.ajax({ 
type: "POST", 
url: "step2.php", 
data: dataString, 
success: function(data) { 
       if(data == 'true') 
        {// your if code 
         } 
      else 
        {// your else code 
        } 

    // This is the button to hide 
    $('.nums').fadeOut('fast'); 
} // end function 
    }); // end ajax   
return false; 
}); 

Answer 3

尝试。

在AJAX success

success: function(data) { 
    if(data == 'false'){ 
     $('.nums').fadeOut('fast'); 
    }else{ 
     // other event if you want 
    } 

在step2.php

echo ($num < 3) ? 'false' : 'true';