这是我的ajax
$(".submit").click(function(){
var vp = $("input#vehicle_plate").val();
var vm = $("input#vehicle_model").val();
var vt = $("input#vehicle_type").val();
var da = $("input#date_acquired").val();
var ad = $("input#assigned_driver").val();
var dataString = 'vehicle_plate='+ vp + '&vehicle_model='+ vm + '&vehicle_type='+ vt + '&date_acquired='+ da + '&assigned_driver='+ ad;
$.ajax({
type: "POST",
url: "process.php",
data: dataString,
success: function(){
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
}
});
return false;
});
这是我的 PHP,我在其中传递“dataString”
<?PHP
include("db.classes.php");
$g = new DB();
$g->connection();
if($_POST)
{
$vehiclePlate = $g->clean($_POST["vehicle_plate"],1);
$vehicleModel = $g->clean($_POST["vehicle_model"],1);
$vehicleType = $g->clean($_POST["vehicle_type"]);
$assignedDriver = $g->clean($_POST["assigned_driver"],1);
$ad = date('Y-m-d', strtotime($_POST["datepicker"]));
$g->add($vehiclePlate, $vehicleModel, $vehicleType, $assignedDriver, $ad);
}
$g->close();
?>
这是我的数据库查询
public function add($vehiclePlate, $vehicleModel, $vehicleType, $assignedDriver, $ad)
{
$sql = "insert into vehicles(`vehicle_plates`,`DA`,`type`, `model`, `driver`) values('$vehiclePlate', '$ad', '$vehicleType', '$vehicleModel', '$assignedDriver')";
if(!mysql_query($sql))
{
$this->error = mysql_error();
return true;
}
else
{
return false;
}
}
AJax 返回成功,但是当我尝试查看数据库中的表时,插入的行都是“未定义”,这似乎是什么原因造成的?
编辑:
这是我的 HTML
<div id="rform">
<form action = "list.php" method="post">
<fieldset>
<legend>Fill Up the Form</legend><br>
<div>
<label class="label-left">*Vehicle Plate:</label>
<input class="label-left" type="text" name="vehicle_plate" id="inputbox1" value maxlength="50">
</div>
<div>
<label class="label-left">*Vehicle Type:</label>
<select class="label-left" id="inputbox" name ="vehicle_type" onchange="document.getElementById('text_content').value=this.options[this.selectedIndex].text">
<option value="Motorcycle">Motorcycle</option>
<option value="Tricycle">Tricycle</option>
<option value="Pick-up">Pick-up</option>
<option value="Truck">Truck</option>
</select><br><br>
</div>
<div>
<label class="label-left">*Vehicle Model:</label>
<input class="label-left" type="text" name="vehicle_model" id="inputbox" value maxlength="50">
</div>
<div>
<label class="label-left">Date Acquired:</label>
<input class="label-left" name="date_acquired" id="datepicker" value maxlength="50"><br><br>
</div>
<div>
<label class="label-left">Assigned Driver:</label>
<input class="label-left" type="text" name="assigned_driver" id="inputbox" value maxlength="50"><br><br>
</div>
<div>
<label class="label-custom" color = red>NOTE: "*" Fields are required</label><br><br>
</div>
<span class="error" style="display:none"> Please Enter Valid Data</span>
<span class="success" style="display:none"> Registration Successfully</span>
<input id="button" type="submit" value="Add" name = "subBtn" class = "submit"/>
</fieldset>
</form>
</div>
最佳答案
问题出在你的选择器上:
var vp = $("input#vehicle_plate").val();
var vm = $("input#vehicle_model").val();
var vt = $("input#vehicle_type").val();
var da = $("input#date_acquired").val();
var ad = $("input#assigned_driver").val();
您的输入选择器没有任何 id
属性,而是需要通过其 name
属性来选择它们(或者您可以提供 id
属性到您的 html 输入元素,以便您的上述选择器可以工作):
var vp = $("input[name=vehicle_plate]").val();
var vm = $("input[name=vehicle_model]").val();
var vt = $("input[name=vehicle_type]").val();
var da = $("input[name=date_acquired]").val();
var ad = $("input[name=assigned_driver]").val();
它会将您的值正确传递到服务器。
关于php - 为什么 Ajax 将 'UNDEFINED' 数据传递给 MySql?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/23322735/