我有一个php页面,用户在文本字段中键入特定的id号并单击“SEARCH”按钮。点击“搜索”后,一个php脚本运行连接到MySQL数据库表“xxx”,并获取与用户输入的id号相匹配的行ID。 SELECT语句为已识别的行抓取数据库值:“productionstage”和“floornotes”。获取选择菜单,以动态显示基于MySQL的特定选择选项选择结果
我需要做的就是把这些结果,并显示他们回来我的窗体页上:
一个选择菜单需要动态显示对应于该行的“productionstage”值,然后一个textarea需要的选项显示“floornotes”的值。
我的代码:
HTML:
<form id="workorderMovement" name='workorderMovement_form' action="workordermovementGET.php" method="post">
<fieldset id="userid">
<span>Welcome <?php echo $user ?> </span>
</fieldset>
<fieldset id="sgnum">
<fieldset id="fieldset" style="text-align: center;">
<span>Please enter the SG Number</span>
</fieldset>
<input type="text" name="sgnumber" id="sgnumber"> <input type="button" name="searchButton" id="searchButton" value="SEARCH">
</fieldset>
<br/>
<br/>
<fieldset id="stageSelectField">
<fieldset id="fieldset" style="text-align: center;">
<span>Please select the Stage Completed</span>
</fieldset>
<select name="stageSelect" id="stageSelect">
<option value="Please Select">Please Select</option>
<option value="Film Done">Film Done</option>
<option value="Staged Done">Staged Done</option>
<option value="Cleanroom Done">Cleanroom Done</option>
<option value="GB2 Done">GB2 Done</option>
<option value="Bagging Done">Bagging Done</option>
<option value="Inspection Done">Inspection Done</option>
<option value="LC Done">LC Inspection Done</option>
<option value="IGU Done">IGU Done</option>
</select>
</fieldset>
<br/>
<br/>
<fieldset id="floorNotesField">
<fieldset id="fieldset" style="text-align: center;">
<span>Please enter any new work order notes</span>
</fieldset>
<textarea type="text" name="floorNotes" id="floorNotes" class="floorNotesText"></textarea>
</fieldset>
<br/>
<br/>
<br/>
</form> <!-- End Work Order Movement Form -->
<fieldset id="doneButtonField">
<input type="button" name="doneButton" id="doneButton" value="DONE">
</fieldset>
MY AJAX:
j("#searchButton").click(function(){
//send Workorder Movement Data values to php using ajax.
var sgnumber = j('#sgnumber').val();
j.ajax ({
method: 'POST',
url: "workordermovementGET.php",
data: {sgNumber: sgnumber},
dataType: 'json',
success: function(data){
if(data.status){
j("select#stageSelect option").filter(function() {
return j(this).val() == data.productionstage;
}).prop('selected', true);
j("textarea#floorNotes").val(data.floornotes);
}
}
});
});
我的PHP:
include('inc.php');
//Get Table Options.
if (isset($_POST['sgNumber'])) {
$sgNumber = $_POST['sgNumber'];
//connect to the database
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if(mysqli_connect_errno()) {
printf('Could not connect: ' . mysqli_connect_error());
exit();
}
$conn->select_db($dbname);
if(! $conn->select_db($dbname)) {
echo 'Could not select database. '.'<BR>';
}
$sql= "SELECT productionstage, floornotes FROM invoices WHERE id = ?";
$stmt = $conn->prepare($sql);
$stmt->bind_param('i', $sgNumber);
$stmt->execute();
$stmt->store_result();
if ($stmt->num_rows == 1) {
$stmt->bind_result($productionstage, $floornotes);
$stmt->fetch();
echo json_encode(array('status' => true, 'productionstage' => $productionstage, 'floornotes' => $floornotes));
} else {
echo json_encode(array('status' => false));
}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//Free the result variable.
$result->free();
//Close the Database connection.
$conn->close();
}//End If statement
?>
我需要知道如何正确地张贴的一部分我的ajax结果作为选择选项(th e数据库值将匹配其中一个预定义的选项),然后取出其余结果并显示在textarea中。另外,如果我可以动态地做到这一点,那将是完美的。
谢谢!
您的代码中有一个SQL注入漏洞! – Hut8
是的,我知道。我正在同时研究如何避免这种情况...任何有关已知链接/教程的建议都非常感谢。谢谢! – rdimouro