我有一个代码,插入和显示记录而不使用ajax和普通的php刷新网页,但我不知道如何使用codeigniter设置它。请帮忙。这是代码
insertion.php
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/
libs/jquery/1.3.0/jquery.min.js">
</script>
<script type="text/javascript" >
$(function() {
$(".comment_button").click(function() {
var test = $("#content").val();
var dataString = 'content='+ test;
if(test=='')
{
alert("Please Enter Some Text");
}
else
{
$("#flash").show();
$("#flash").fadeIn(400).html('<img src="ajax-loader.gif" align="absmiddle">
<span class="loading">Loading Comment...</span>');
$.ajax({
type: "POST",
url: "demo_insert.php",
data: dataString,
cache: false,
success: function(html){
$("#display").after(html);
document.getElementById('content').value='';
document.getElementById('content').focus();
$("#flash").hide();
}
});
} return false;
});
});
</script>
// HTML code
<div>
<form method="post" name="form" action="">
<h3>What are you doing?</h3>
<textarea cols="30" rows="2" name="content" id="content" maxlength="145" >
</textarea><br />
<input type="submit" value="Update" name="submit" class="comment_button"/>
</form>
</div>
<div id="flash"></div>
<div id="display"></div>
demo_insert.php PHP代码显示最近从数据库中插入的记录。
<?php
include('db.php');
if(isSet($_POST['content']))
{
$content=$_POST['content'];
mysql_query("insert into messages(msg) values ('$content')");
$sql_in= mysql_query("SELECT msg,msg_id FROM messages order by msg_id desc");
$r=mysql_fetch_array($sql_in);
}
?>
<b><?php echo $r['msg']; ?></b>
答案 0 :(得分:0)
在 wellcome 控制器中,您可以添加以下内容:
public function inserting()
{
$this->load->view('inserting');
}
public function process()
{
$content=$this->input->post('content');
if($this->db->insert('mytable', array('msg' => $content))){
echo "<b>{$content}</b>";
}
然后,您应该在应用程序/视图中使用insertion.php作为您的视图,并且ajax网址将是 / process 。
没有测试过,但这应该可以解决问题。另外,您应该查看此示例http://runnable.com/UXczcazDrMMiAAGl/how-to-do-ajax-in-codeigniter-for-php