仅当我的数据变量不为空时才如何关注表单字段 [英] How to focus on a form field only if my data variable is not empty

问题描述

我一直在使用php和ajax来验证数据库中是否存在以表单形式插入的电子邮件.

I have been using php and ajax to validate if an email inserted in my form exists in my database.

我正在使用jquery将电子邮件值发送到我的php文件，如果找到了电子邮件，则返回一条消息.我的代码工作正常，但是我希望如果找到电子邮件，则光标将集中在#usu_email字段上，直到更改电子邮件为止.之后，它应该允许我继续下一个领域.

I am using jquery to send the email value to my php file and return a message if the email is found. My code is working fine but I want if an email is found the cursor be on focus on the #usu_email field until the email be changed. After this, it should allow me to continue to next field.

这是我正在使用的jquery代码:

This is the jquery code I am using:

function getemail(value) {
  var usumail = $("#usu_email").val();

  $.ajax({
    type: "POST",
    url: "ajax_email.php",
    data: "usu_email=" + usumail,
    success: function(data, textStatus) {
      if (data !== null) {
        $("#eresult").html(data);
        $("#usu_email").focus();
      }
    },
  });
};

我的问题是，如果数据库中不存在和电子邮件，则光标将继续关注#usu_email字段，并且不允许我继续下一个字段.

My problem is that if and email does not exist in my database the cursor keeps doing focus on my #usu_email field and does not allow me to continue to next field.

对于这个问题，我将不胜感激，因为我对jquery的了解很少.

I will appreciate any help about this problem because I know very little about jquery.

推荐答案

首先...您的条件if (data !== null) 始终将为true，因为将始终提供数据.一个空字符串.

First... Your condition if (data !== null) always will be true since there always will be a data provided... Be it an empty string.

唯一不会出现data的情况是Ajax错误...并且该条件甚至不会被评估，因为success回调将不会执行.

The only case where there will be no data is on Ajax error... And the condition won't even be evaluated because the success callback won't execute.

接下来，我假设您的Ajax请求是在$("#usu_email")模糊条件下触发的...否则，我不知道您如何实现«不允许我继续».

Next, I assume that your Ajax request is triggered on $("#usu_email") blur... Else, I don't know how you achieve «does not allow me to continue».

以这种方式修改它以比较响应:

Modify it in this way to compare a response:

function getemail(value) {
  var usumail = $("#usu_email").val();

  $.ajax({
    type: "POST",
    url: "ajax_email.php",
    data: "usu_email=" + usumail,
    datatype: "json",
    success: function(data) {  // There is only one argument here.
      // Display the result message
      $("#eresult").html(data.message);

      if (data.email_exist == "yes") {
        $("#usu_email").focus();
      }
      if (data.email_exist == "no") {
        // Something else to do in this case, like focussing the next field.
      }
    },
  });
};

在PHP方面，您必须提供json响应.看起来像这样:

On the PHP side, you have to provide the json response. It would look like something like this:

<?php

// You have this variable to compare against the database
$email = $_POST[usu_email];

// You say it is working.
// ...
// Then, you certainly have a result... Say it's $found (true/false).

// Build an array of all the response param you want to send as a response.
if($found){
  $result[email_exist] = "yes";
  $result[message] = "The submitted email already exist.";
}else{
  $result[email_exist] = "no";
  $result[message] = "A success message about the email here.";
}

// Add this header to the returned document to make it a valid json that doesn't need to be parsed by the client-side.
header("Content-type:application/json");

// Encode the array as a json and print it. That's what is sent in data as an Ajax response.
echo json_encode($result);

?>

请注意不要回声其他任何内容.甚至没有空格或行返回.

Be carefull not to echo anything else. Not even a blank space or a line return.

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