1

我正在使用 jquery/ajax 提交表单而不刷新页面。

这是我的表单代码：

<div id="event_form">
    <form name="event_submit" method="post" action="">
        <label for="what" id="what_label">What is the event?</label>
        <input type="text" name="what" id="what" class="entry_field" />
        <br />
        <label class="error" for="what" id="what_error">This field is required.</label>
        <br />
        <label for="where" id="where_label">Where is the event?</label>
        <input type="text" name="where" id="where" class="entry_field" />
        <br />
        <label class="error" for="where" id="where_error">This field is required.</label>
        <br />
        <label for="when" id="when_label">When is the event?</label>
        <input type="text" name="when" id="when" class="entry_field" />
        <br />
        <label class="error" for="when" id="when_error">This field is required.</label>
        <br />
        <label for="details" id="details_label">Additional details</label>
        <textarea rows="8" cols="30" name="details"></textarea>            
        <br />
        <input type="submit" value="Post Event" name="submit" class="submitButton" title="Click to post event" />
    </form>
</div>

这是我的 jquery/ajax：

  $(function() {
    $('.error').hide();

    $(".submitButton").click(function() {
      // validate and process form
      // first hide any error messages
      $('.error').hide();

      var what = $("input#what").val();
      if (what == "") {
        $("label#what_error").show();
        $("input#what").focus();
        return false;
      }
      var where = $("input#where").val();
      if (where == "") {
        $("label#where_error").show();
        $("input#where").focus();
        return false;
      }
      var when = $("input#when").val();
      if (when == "") {
        $("label#when_error").show();
        $("input#when").focus();
        return false;
      }

      var dataString = 'what='+ what + '&where=' + where + '&when=' + when;
      //alert (dataString);return false;

      $.ajax({
        type: "POST",
        url: "event_submit.php",
        data: dataString,
        success: function() {
          $('#event_form').html("<div id='message'></div>");
          $('#message').html("<h2>Event Submitted!</h2>");
        }
       });
      return false;
    });
  });

这是php：

<?php

include "conf.php";

if ((isset($_POST['what'])) && (strlen(trim($_POST['what'])) > 0)) {
    $what = stripslashes(strip_tags($_POST['what']));
} else {$what = 'No description of the event entered.';}
if ((isset($_POST['where'])) && (strlen(trim($_POST['where'])) > 0)) {
    $where = stripslashes(strip_tags($_POST['where']));
} else {$where = 'No location entered for the event.';}
if ((isset($_POST['when'])) && (strlen(trim($_POST['when'])) > 0)) {
    $when = stripslashes(strip_tags($_POST['when']));
} else {$when = 'No time entered for the event.';}
if ((isset($_POST['details'])) && (strlen(trim($_POST['details'])) > 0)) {
    $details = stripslashes(strip_tags($_POST['details']));
}

if(isset($details)){
    mysql_query("INSERT INTO `events` (`school_id`, `what`, `where`, `when`, `details`) VALUES ('$school_id', '$what', '$where', '$when', '$details')") or die(mysql_error());
}
else {
    mysql_query("INSERT INTO `events` (`school_id`, `what`, `where`, `when`) VALUES ('$school_id', '$what', '$where', '$when')") or die(mysql_error());
}

exit;
?>

在表单所在的页面上，URL 中有一个“$school_id”，如 ?school=1687 或其他。我可以使用 $_GET 在表单页面上检索它，但是如何将此变量传递给 ajax/php？我需要将它作为变量插入到表中。

4

4 回答 4

1

怎么样的东西

<input type="hidden" name="schoolid" value="<?php echo $_GET['school_id']; ?>">

以你的形式？请理解，回显未清理的 GET 参数仅用于说明目的 - 它使您对 CSS 敞开心扉！

于 2012-11-27T00:14:53.383 回答
0

您可以只使用隐藏的输入。就像是：

<input type="hidden" id="school_id" value="<?php echo $school_id" />

然后将其附加到您的数据变量上：

var dataString = 'school_id='+$('#school_id').val()+'&what='+ what + '&where=' + where + '&when=' + when;

然后在您的 php 页面上检索它：

$school_id = $_POST['school_id'];
于 2012-11-27T00:15:00.120 回答
0

尝试：

//Retrieving $_GET['school_id']  and passing as JSON object
var school_id = <?php echo $_GET['school_id']; ?>
var dataString = { 'what': what, 'where' : where, 'when': when, 'school_id': school_id }
$.ajax({
    type: "POST",
    url: "event_submit.php",
    data: dataString,
    type: json,
    success: function() {
      $('#event_form').html("<div id='message'></div>");
      $('#message').html("<h2>Event Submitted!</h2>");
    }
   });

$_POST['school_id']您可以在 上访问它event_submit.php。

于 2012-11-27T01:14:49.803 回答
0

在表单中添加一个隐藏字段，如下所示

 <input type="hidden" id="school_id" value="<?php echo $_REQUEST['school']; ?>" />

现在在 jQuery 脚本中，您可以通过一行代码简单地获取所有表单元素值

var dataString = $('#form-id-here').serialize();

然后通过ajax发送

$.ajax({
    type: "POST",
    url: "event_submit.php",
    data: dataString,
    success: function() {
      $('#event_form').html("<div id='message'></div>");
      $('#message').html("<h2>Event Submitted!</h2>");
    }
   });

最后将mysql插入查询更改为followig

   mysql_query("INSERT INTO `events` (`school_id`, `what`, `where`, `when`, `details`) VALUES ('".$_POST['school_id']."', '$what', '$where', '$when', '$details')") or die(mysql_error());

我想现在你已经拥有了所有需要的东西。享受 ：）

于 2012-11-27T05:28:56.220 回答