我看起来非常努力,但我无法理解使用PHP的AJAX。
这就是我所拥有的,当用户点击下拉列表时,我希望它保存到我的数据库中
<select>
<?php $taskStatus = "SELECT * FROM task_status WHERE used = 1 ORDER BY id ASC ";
$taskresults = $conn->query($taskStatus) or die(mysqli_error($conn));
while($taskStatusRow = mysqli_fetch_assoc($taskresults)) {
echo " <option value= ". $taskStatusRow['name'] ." >". $taskStatusRow['name'] ." </option>";
}
?>
</select>
这是我想要运行的查询:
INSERT INTO snagging (taskstatus, updated_at)
WHERE ID = 1234
VALUES taskStatusRow['name'], $now);
答案 0 :(得分:1)
我将在这里为您提供整体AJAX流程。我试图提供评论以显示控制流程。
12
jQuery + AJAX
nov_dec
在<select id="selectOption"> //******* Assign an ID
<?php $taskStatus = "SELECT * FROM task_status WHERE used = 1 ORDER BY id ASC ";
$taskresults = $conn->query($taskStatus) or die(mysqli_error($conn));
while($taskStatusRow = mysqli_fetch_assoc($taskresults)) {
echo " <option value= ". $taskStatusRow['name'] ." >". $taskStatusRow['name'] ." </option>";
}
?>
</select>
内,您可以访问通过AJAX传递的数据,如:
$(document).ready(function() {
$("#selectOption").change(function(){ //** on selecting an option based on ID you assigned
var optionVal = $("#selectOption option:selected").val(); //** get the selected option's value
$.ajax({
type: "POST", //**how data is send
url: "MYPROCESSPAGE.php", //** where to send the option data so that it can be saved in DB
data: {optionVal: optionVal }, //** send the selected option's value to above page
dataType: "json",
success: function(data){
//** what should do after value is saved to DB and returned from above URL page.
}
});
});
});
希望它有所帮助。