我想将输入字段中的数据保存到mysql数据库中,因此首先创建一个模式窗口和输入字段:
<!-- Button trigger modal -->
<button class="btn btn-success" data-toggle="modal" data-target="#myModal">
Add new</button>
<div id="output"></div>
<!-- Modal -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
<h4 class="modal-title" id="myModalLabel">Add new row</h4>
</div>
<div class="modal-body">
......
<div class="input-group">
<span class="input-group-addon">Ime</span>
<input type="text" id="Ime" class="form-control" placeholder="Upisi ime">
</div>
</br>
<div class="input-group">
<span class="input-group-addon">Pol</span>
<input type="text" id="pol" class="form-control" placeholder="Pol (male/female)">
</div>
</br>
<div class="input-group">
<span class="input-group-addon">Godine</span>
<input type="text" id="godine" class="form-control" placeholder="Godine starosti">
</div>
</br>
<div class="input-group">
<span class="input-group-addon">Broj pojedenih krofni</span>
<input type="text" id="krofne" class="form-control" placeholder="Pojedene krofne">
</div>
</br>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button id="newData" class="btn btn-primary">Add new data</button>
</div>
</div>
<!-- /.modal-content -->
</div>
<!-- /.modal-dialog -->
</div>
<!-- /.modal -->
现在我编写jQuery AJAX代码来向数据库添加数据:
<script>
//add data to database using jquery ajax
$("#newData").click(function() {
//in here we can do the ajax after validating the field isn't empty.
if($("#ime").val()!="") {
$.ajax({
url: "add.php",
type: "POST",
async: true,
data: { Name:$("#ime").val(), Gender:$("#pol").val(), Age:$("#godine").val(), Donuts_eaten:$("#krofne").val()}, //your form data to post goes here as a json object
dataType: "html",
success: function(data) {
$('#output').html(data);
drawVisualization();
},
});
} else {
//notify the user they need to enter data
}
});
</script>
最后我创建了一个php文件(add.php)
<?php
$con = mysql_connect('localhost', 'gmaestro_agro', 'pass') or die('Error connecting to server');
mysql_select_db('gmaestro_agro', $con);
mysql_select_db('gmaestro_agro', $con);
$query = "INSERT INTO `stat` (`Name`, `Gender`, `Age`, `Donuts eaten`) VALUES (";
$query .= mysql_real_escape_string($_POST['Name']) . ", ";
$query .= mysql_real_escape_string($_POST['Gender']) . ", ";
$query .= mysql_real_escape_string($_POST['Age']) . ", ";
$query .= mysql_real_escape_string($_POST['Donuts_eaten']);
$query .= ")";
$result = mysql_query($query);
if($result != false) {
echo "success!";
} else {
echo "an error occured saving your data!";
}
?>
现在,当我试图添加数据时,我得到了一个错误:
an error occurred saving your data!
。这里到底有什么问题?我一整天都在试图找出错误。。。
最佳答案
您没有引用字符串:
$query .= mysql_real_escape_string($_POST['Name']) . ", ";
应该是:
$query .= "'" . mysql_real_escape_string($_POST['Name']) . "', ";
(对于所有字符串值)
顺便说一下,如果您切换到PDO或mysqli并准备好语句,可能会让您的生活更轻松。那么您就不必转义和引用变量,而且
mysql_*
函数也不推荐使用。关于javascript - jQuery ajax添加到mysql数据库,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/21499192/