Question

我正在尝试创建一个弹出模式，它将显示从php提交脚本回显的文本。到目前为止它的成功并打印出回显的内容，但我也希望在php脚本中发生异常时捕获回显的消息。

popnotify是一个单独的js函数，它接受文本输入并以文本作为段落显示模态。

createuser.php

try:
    ......
    echo 'Successfully created a new user';
}
catch (PDOException $ex){
    echo "Something went wrong when creating a new user";
    http_response_code(403);
    die();
}

的.js

jQuery("#load_form").on("click","#create_submit",function(event){
event.preventDefault();
var to_submit = $("#s_form").serialize();
var formLink = $("#s_form").attr('action');
jQuery('#load_form').find('input, select, button').prop('disabled', true);

$.post(formLink,to_submit)
.done(function(data){
    popnotify_success(data);
})
.fail(function(data){ 
    popnotify_fail(data);
})
.always(function(){
    jQuery('#load_form').find('input, select, button').prop('disabled', false);  
});  });

Answer 1

如何使用标志来确定是否存在错误

您的PHP，createuser.php

try {
    // Some more code here that tries to create the user
    $response = [
        'error' => 0,
        'message' => 'Successfully created a new user',
    ];
    echo json_encode($response);
}
catch (PDOException $ex){
    $response = [
        'error' => 1,
        'message' => 'Something went wrong when creating a new user',
    ];
    echo json_encode($response);
    // HTTP Code 200 OK, there were not HTTP errors or server (500) errors
    // http_response_code(403);
    die();
}

您的JavaScript

jQuery("#load_form").on("click","#create_submit",function(event){
  event.preventDefault();
  var to_submit = $("#s_form").serialize();
  var formLink = $("#s_form").attr('action');
  jQuery('#load_form').find('input, select, button').prop('disabled', true);

  $.post(formLink,to_submit)
  .done(function(data){
    // Parse the result into an object
    var result = JSON.parse(data);

    // Get the error flag from the object
    if (result.error == 0) {

      // There was no error
      popnotify_success(result.message);

    } else {

      // There was an error
      popnotify_fail(result.message);

    }
  })
  .fail(function(data){ 
    // Handle http errors here
  })
  .always(function(){
    jQuery('#load_form').find('input, select, button').prop('disabled', false);  
  });  
});

如果所有函数都创建通知，您也可以使用通用popnotify()函数，而不是两个函数。依赖于错误/成功的任何其他代码都可以继承处理函数。

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1 个答案: