Php jqueryajax Post,将它们添加到DB中,返回数据并再次重复该过程
我完全糊涂了,所以请让我把逻辑解释清楚 这里是我从数据库获取数据的地方Php jqueryajax Post,将它们添加到DB中,返回数据并再次重复该过程,php,mysql,jquery,post,Php,Mysql,Jquery,Post,我完全糊涂了,所以请让我把逻辑解释清楚 这里是我从数据库获取数据的地方 index.php $sorular_hepsi = mysql_query("select * from tblsorular where hafta=1 order by rand() limit 2"); $soru_ust = mysql_fetch_assoc($sorular_hepsi); $soru_id = $soru_ust_rs["id"]; $soru_grup_i
index.php
$sorular_hepsi = mysql_query("select * from tblsorular where hafta=1 order by rand() limit 2");
$soru_ust = mysql_fetch_assoc($sorular_hepsi);
$soru_id = $soru_ust_rs["id"];
$soru_grup_id = $soru_ust["sId"];
$soru1 = $soru_ust["soru"];
$sorular = mysql_query("select * from tblsorular where sId=$soru_grup_id");
$totalKayit = mysql_num_rows($sorular_rs);
while ( $sorular_rs=mysql_fetch_assoc($sorular)) {
$sorular2[] = $sorular_rs["soru"];
$sorular2Id[] = $sorular_rs["id"];
}
$userId = 1234;
下面是index.php
<div id="sorugonder" class="soruStyle">
<a href="#" class="sorugonder" id="<?=$sorular2Id[0]?>"><?=$sorular2[0]?></a>
</div>
<div id="soru_sag" class="soruStyle">
<a href="#" class="sorugonder2" id="<?=$sorular2Id[1]?>"><?=$sorular2[1]?></a>
</div>
$(function() {
$(".sorugonder").click(function() {
// $('#load').fadeIn();
var commentContainer = $(this).parent();
var id = $(this).attr("id");
var string = 'id='+ id ;
$.ajax({
type: "POST",
url: "islem.php?islem=soruKayit",
data: string,
cache: false,
success: function(data){
commentContainer.slideUp('slow', function() {$(this).remove();});
$('#sorugonder').fadeOut(1000);
$('#soru_sag').fadeOut(2000);
console.log(string);
// alert(id);
}
});
return false;
});
});
此功能将控制台日志打印为id=xx
这是我的islem.php页面,用于获取ajax数据
if(isset($_POST["islem"]) && $_POST["islem"]=="soruKayit"){
header('Content-type: application/json');
$id= $_POST['string'];
$cevapTarihi = date("d-m-Y H:i:s");
$cevapId = json_encode($id);
$userId = 1234;
$kayit = @mysql_query("INSERT INTO tblk_skor VALUES(NULL, $userId,$cevapId,$cevapTarihi)");
if(!$kayit){
echo "Error:".mysql_error();
}
die();
}
我有4张不同的png图片,我用它们作为背景。例如:
$1 = '<img src="../img/1.png" />';
$2 = '<img src="../img/2.png" />';
$3 = '<img src="../img/3.png" />';
$4 = '<img src="../img/4.png" />';
$1='';
$2 = '';
$3 = '';
$4 = '';
我已经在每个后期处理中更改了我列出的数据背景。但我不知道该怎么放,放在哪里
$.ajax函数不会将数据发送到islem.php,否则无法获取数据
简单地说:
有什么建议吗?ajax中的
data
选项应该包含要发送到服务器的数据
$.ajax({
type: "POST",
url: "islem.php",
data: [{ name:'islem', value:'soruKayit' },
{ name:'id', value:id }],
cache: false,
....