我想用字符的 id 更新我的数据库,但是当我将它们放入插槽时,它不会更新我希望它更新的行。我的问题是你能指出我如何正确编码或修复我的错误的正确方向吗?
function updateTeam(){
var team = '', slot = [];
if($('input[name=s0]').val()!=""){
slot.push($('input[name=s0]').val());
}
if($('input[name=s1]').val()!=""){
slot.push($('input[name=s1]').val());
}
if($('input[name=s2]').val()!=""){
slot.push($('input[name=s2]').val());
}
$.each(slot, function(i,e){
if(i == 0) team = e;
else team = team + ',' + e;
});
$.ajax({
url : _path + "/core/ajax.php",
type : 'POST',
data : { f: 'setTeam', i: team},
dataType : 'json',
success : function(data) {
if(data.error){
errorMessage('Error: ' + data.error, data.error, data.error);
}
}
});
}
php
function clean($content) {
$content = mysql_real_escape_string(htmlspecialchars($content));
return $content;
}
//Update the user team.
if (isset($_POST['f']) && $_POST['f'] == 'updateTeam') {
if (isset($_POST['s0'])) {
$cid1 = $secure->clean($_POST['s0']);
} else {
$cid1 = '1';
}
if (isset($_POST['s1'])) {
$cid2 = $secure->clean($_POST['s1']);
} else {
$cid2 = '2';
}
if (isset($_POST['s2'])) {
$cid1 = $secure->clean($_POST['s2']);
} else {
$cid1 = '3';
}
$updateTeam = $db->query("UPDATE accounts SET cid1 = '$cid1', cid2 = '$cid2', cid3 = '$cid3' WHERE id = '$id'");
}
当我用谷歌浏览器检查元素时,它显示 i:1,5,2。显示我将如何更新我的行,以便“i”中的 1=$cid1、5=$cid2 和 2=cid3,是我的 php 代码错误吗?html:
<div id="droppable_slots" class="current_team">
<div class="slot 1">1</div>
<input type="hidden" name="s0" value="10">
<div class="slot 2">2</div>
<input type="hidden" name="s1" value="7">
<div class="slot 3">3</div>
<input type="hidden" name="s2" value="3">
</div>