我是AJAX的新手,所以如果解决方案显而易见,我很抱歉。我必须在不刷新页面的情况下将日期发送到MySQL表,而我目前在这方面遇到了问题。
代码给出我遇到以下错误:
"注意:未定义的索引:张贴在inserisci.php第6行"
FORM
<form id="form" class="form-horizontal shadow-z-1" action="includes/inserisci.php" method="post">
<fieldset>
<legend style="text-align: center; position: relative; top: 8px;">Inserisci un Post</legend>
<hr>
<div class="form-group">
<label for="inputTitle" class="col-lg-2 control-label">Titolo</label>
<div class="col-lg-10">
<input type="text" class="form-control" id="inputText" name="titolo" placeholder="Scrivi qui il Titolo">
</div>
</div>
<div class="form-group">
<label for="textArea" class="col-lg-2 control-label" name="testo">Messaggio</label>
<div class="col-lg-10">
<textarea id="messaggio"></textarea>
</div>
</div>
<div class="form-group is-empty is-fileinput">
<label for="inputFile" class="col-md-2 control-label">File</label>
<div class="col-md-10">
<input type="text" readonly="" class="form-control" placeholder="Browse..." pmbx_context="19E61A0C-3526-4E51-8535-935982C4C335">
<input type="file" id="inputFile" multiple="" pmbx_context="1D2BCAEA-08CC-476A-8F4A-EF6BD51B9102">
</div>
<span class="material-input"></span></div>
<div class="col-md-10 col-md-offset-2">
<button type="button" class="btn btn-default" onclick="document.getElementById('modalposta').style.display = 'none';">Cancel</button>
<button type="submit" class="btn btn-primary" onClick='send(); return false;' >Submit</button>
</div>
</fieldset>
</form>
INSERISCI.JS
function send(){
var content = $.ajax({
type: "POST",
url: "index.php",
data: {titolo:titolo & post:testo}
})
.done(function() {
alert( "success" );
})
.fail(function() {
alert( "error" );
})
.always(function() {
alert( "complete" );
});
}
INSERISCI.PHP
<?php
include('../core.php');
$titolo=$_REQUEST ['titolo'];
$post=$_REQUEST ['post'];
$sql = mysql_query("");
mysql_close();
?>
其他问题。我需要在不刷新页面的情况下重新加载此内容。
索引.PHP
<?php $sql = mysql_query("") or die ("Nessun errore");
if (mysql_num_rows($sql) > 0)
{
while ($row = mysql_fetch_assoc($sql))
{
echo '
<article class="flipper white-panel">
<a class="flipcard flip" style="
position: absolute;
right: 0px;
top: -10px;
z-index:1;
"><i class="material-icons" style="font-size: 40px;">bookmark</i></a>
<div class="articolo">
<img src="http://i.imgur.com/sDLIAZD.png" alt="">
<h4><a href="#">'.$row ['titolo'].'</a></h4>
<p>'.Markdown($row ['contenuto']).'</p>
<div class="pull-right">
<span class="label label-default">alice</span>
<span class="label label-primary">story</span>
<span class="label label-success">blog</span>
<span class="label label-info">personal</span>
<span class="label label-warning">Warning</span>
<span class="label label-danger">Danger</span>
</div>
<hr>
</div>
<div class="commenti">
<ul class="comment-list">
<li>
<div class="comment-img">
</div>
<div class="comment-text">
</div>
</li>
</ul>
<input type="text" class="form-control inputcomment" placeholder="Lascia un Commento.."/>
</div>
</article>
';
}
}
谢谢大家。
注意:
- 请不要使用
mysql_*
函数。它们已被弃用- 您的代码容易受到SQL注入攻击。请务必向他们致辞
这是一个语法错误。这不是一个有效的JavaScript对象。将data
对象更改为:
data: {titolo:titolo, post:testo}
此外,在使用.done
之前,您需要关闭$.ajax
。
var content = $.ajax({
type: "POST",
url: "index.php",
data: {titolo:titolo & post:testo} // Should be closed here.
})
.done(function() {
alert( "success" );
}) // Should not be closed. Remove ;
.fail(function() {
alert( "error" );
}) // Should not be closed. Remove ;
.always(function() {
alert( "complete" );
});
// }); Remove this!
如果没有结果,请更改这些行:
$titolo = $_POST['titolo'];
$post = $_POST['post'];
等等,你得到这两个的值了吗?
var titolo = $('input[name="titilo"]').val();
var testo = $('input[name="testo"]').val();
这是因为您在AJAX请求中传递的数据不是有效的javascript对象,并且您没有从表单中获取变量:
data: {titolo:titolo & post:testo}
你应该做一些类似的事情:
var titolo = $('form input[name="titolo"]').val();
var testo = $('form input[name="testo"]').val();
data: {
titolo:titolo,
post:testo
}
此外,顺便说一句,当访问数组元素时,在变量和括号之间放一个空格是不好的。我很惊讶这竟然有效。
$_REQUEST ['titolo']
应该是
$_REQUEST['titolo']
如果您想使用ajax表单提交,请使用更新表单定义
<form id="form" class="form-horizontal shadow-z-1" action="#" method="post">
使用以下更新的javascript
function send() {
var content = $.ajax({
type: "POST",
url: "process.php",
data: {
titolo: "Value for titolo",
post: "Value for post"
}
}).done(function() {
alert( "success" );
}).fail(function() {
alert( "error" );
}).always(function() {
alert( "complete" );
});
}