Javascript PHP MySQL清单数据问题
我有一些列表数据问题 数据库架构: 这是我的密码: SQL查询:Javascript PHP MySQL清单数据问题,javascript,php,mysql,ajax,database,Javascript,Php,Mysql,Ajax,Database,我有一些列表数据问题 数据库架构: 这是我的密码: SQL查询: SELECT tblquestions.Question, tblhints.hint, tblhints.pic, tblquestions.Id FROM tblquestions INNER JOIN tblhints ON tblquestions.GroupId = tblhints.GroupId ORDER BY RAND() LIMIT 57 它给出如下输出: 在TBL问题中,字
SELECT tblquestions.Question,
tblhints.hint,
tblhints.pic,
tblquestions.Id
FROM tblquestions INNER JOIN tblhints ON tblquestions.GroupId = tblhints.GroupId ORDER BY RAND() LIMIT 57
它给出如下输出:
在TBL问题中,字段记录如下:
Php代码:
<?php
$i=0;
$t=0;
for ($i=0; $i < 57 ; $i++) {
while($kayitlar_rs = mysql_fetch_array($kayitlar)) {
// $toplamkayit = mysql_num_rows($kayitlar);
$soruGroupId = $kayitlar_rs["GroupId"];
$hint = $kayitlar_rs["hint"];
$sorufoto = $kayitlar_rs["pic"];
// $sorubtntxt = array();
$sorusql = mysql_query("select * from tblquestions where GroupId=".$soruGroupId);
while($soruRs = mysql_fetch_array($sorusql)) {
$sorubtntxt[] = $kayitlar_rs["Question"];
}
// die();
$i = $i+1;
$t = $t+1;
?>
HTML/DIVs:
<div id="soru<?=$i?>">
<span class="hint hintbg hintyazi">İPUCU: <b class="orange"><?=$hint?></b></span>
<div class="sorufoto"><img src="../assets/ulke/<?=$sorufoto?>"></div>
<span class="soruId"><?=$soruId?></span>
<div id="sorubtn" class="font">
<div id="soruBtnSol"><a class="sorubtn<?=$i?>" href="#"><span class="sorubtnTxtSol"><?=$sorubtntxt[0]?></span></a></div>
<div id="soruBtnSag"><a class="sorubtn<?=$t?>" href="#"><span class="sorubtnTxtRight"><?=$sorubtntxt[1]?></span></a></div>
</div>
</div>
有人有合作伙伴吗。代码引用了$kayitlar\r[“GroupId”]
,但GroupId不在查询的选择列表中。2.记录被提取到$soruRs
中,然后这个变量就永远不会被使用。3.不清楚为什么$i
会增加两次$t
是$i/2
那么拥有两个变量有什么意义呢?