我正在制作这个Flag/Unflag
系统,它工作正常。我唯一要苦苦挣扎的事情是单击后如何更新Flag_icon
?应该是这样,它在单击后才更改,如果再次单击,则它会变回来。现在,我必须单击该标志,然后手动重新加载页面,然后对其进行更改。
$FlagStatus[$count]
在我的数据库中的值为YES/NO
,而$testjobids[$count]
是数据库表中的唯一ID。我一直在寻找一些Ajax和Javascript来做到这一点,但是我似乎无法正确地实现它。我只需要指出正确的方向,因为我有点被卡住了。
Index.php
if ($FlagStatus[$count] == ('YES')) {
echo'<img class="Unflagging" onclick="changeImage('.$testjobids[$count].')" id="'.$testjobids[$count].'" data-id = "'.$testjobids[$count].'" src = "../Test/Images/Flag/FlagMarked.png">';
} elseif($FlagStatus[$count] == ('NO')){
echo'<img class="Flagging" onclick="changeImage()" id="'.$testjobids[$count].'" data-id2 = "'.$testjobids[$count].'" src = "../Test/Images/Flag/FlagUnmarked.png">';
}
Flagging.php /与Unflagging.php几乎相同
<?php
require("../resources/MysqliHandler.class.php");
require("../resources/layoutfunctions.php");
require("GetData.php");
$ICON_DIMENSION = "16px";
// Used to connect the right server Testreportingdebug or Testreporting
$db = ServerConn();
// -------------------------------------
$MysqliHandler = new mysqliHandler($db);
$MysqliHandler->query("SET NAMES UTF8");
$receiver = $_POST["FlagID"];
$MarkYes = "UPDATE `testreportingdebug`.`testjob` SET `FlagStatus` = 'YES' WHERE id = $receiver";
$query = $MysqliHandler->query($MarkYes);
?>
Ajax
echo'
<script type="text/javascript">
$(document).on("click", ".Unflagging", function(){
var FlagID = $(this).data("id");
$.ajax({
method: "post",
url: "unflagging.php",
data: { FlagID: FlagID},
success: function(data) {
changeImage();
}
});
});
$(document).on("click", ".Flagging", function(){
var FlagID = $(this).data("id2");
$.ajax({
method: "post",
url: "flagging.php",
data: { FlagID: FlagID},
success: function(data) {
changeImage();
}
});
});
</script>
';