在Ajax jQuery中显示多个错误结果

我试图同时显示多个错误结果（如果有一个以上的错误结果），而不是逐个显示。让用户一次看到全部错误，以便一次修复。

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这是我的代码：在主页

<form id="iform" name="iform" onsubmit="validate(); return false">
<table width="509"><tr>
<td width="114">Name User</td>
<td width="177"><input type="text" name="owner" id="owner"/></td><td id="err1" width="202" style="color:red;"></td></tr><tr>
<td>Email </td><td><input type="email" name="email" id="email"/></td><td id="err2" style="color:red;"></td></tr><tr>
<td>Paypal</td>
<td><input type="email" name="paypal" id="paypal" /></td><td id="err3" style="color:red;"></td></tr><tr>
<td>&nbsp;</td>
<td><input type="submit"  value="Submit" /></td></tr>
</table>
</form>

Ajax

$.ajax({
    url:'engine.php',
    data:{
    action:'add_user',  
    owner:document.iform.owner.value,
    email:document.iform.email.value,
    paypal:document.iform.paypal.value
    },
    type:'POST',
    success: function(data){
        if (data=="er1"){
        $("#err2").html("This email is already existed, please try another");
        return false;
        }
        if (data=="er2"){
        $("#err3").html("This paypal account is already existed, please try another");
        return false;
        }
        if (data=="ok"){
        location.reload();
        }
    }
    });

在PHP中

$owner=$_POST[owner];
$email=$_POST[email];
$paypal=$_POST[paypal];
  $ss = $conn->prepare('select email_owner, paypal_owner from tb_owner');
  $ss->execute();
  $watashi=$ss->fetch();
if ($email==$watashi['email_owner']){
    echo"er1";
}
else if ($paypal==$watashi['paypal_owner']){
    echo"er2";
}
else{
/* EVEYTHING IS OK ->INSERT TO DATABASE */
 if ($stmt->rowCount()){ 
echo"ok";
}
}

$errors = array();
if($email==$watashi['email_owner']){
    $errors[] = "this email is already existed, try another";
}
if($paypal==$watashi['paypal_owner']){
    $errors[] = "this paypal is already existed, try another";
}
if(count($errors) > 0){
    echo json_encode(array('status' => 'error', 'errors' => $errors));

success: function(data){
    var json = $.parseJSON(data);

       if(json.status=== 'error'){
        $("#err2").html(json.errors[0]);
        $("#err3").html(json.errors[1]);
        return false;
         }

        if(json.status==="success"){
        location.reload();
        }