【发布时间】:2015-06-12 00:13:03
【问题描述】:
首先我从 MYSQl 获取数组,当 Ajax 显示更改值结果时我选择输入
echo "<select onchange='showUser(this.value)'> <option>Select Unit</option>";
$query="select * from item_units WHERE item_id='".$item_id."'";
$res=mysql_query($query);
if(!$res){die(mysql_error());}
while($row=mysql_fetch_array($res)){
$item_unit =$row['item_unit'];
$query4 = "SELECT unit_name FROM unit WHERE unit_id='".$item_unit."'";
$rest4 = mysql_query($query4);
if(isset($rest4))
$unit_name4 = mysql_result($rest4, 0);
echo "<option value='$item_unit'>$unit_name4</option>";
}
echo"</select><br /> ";
然后我从 mysql 获取 $item_id
$query="select * from items WHERE item_id='".$item_id2."'";
$res=mysql_query($query);
if(!$res){die(mysql_error());}
while($row=mysql_fetch_array($res)){
$item_id=$row['item_id'];}
然后使用
echo"<script>
function showUser(str) {
if (str == '') {
document.getElementById('txtHint').innerHTML = '';
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById('txtHint').innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open('GET','myesso.php?id=$item_id&q='+str,true);
xmlhttp.send();
}
} ";
但是 ID = $item_id 的值没有任何值,因为 OF 超过了每个记录的值
我如何解决它我可以将它作为数组发送吗?
Myesso.php
Myesso.php
<?php
session_start();
include "config.php";
$q = intval($_GET['q']);
$gid = $_REQUEST['id'];
$moha = $_SESSION['logid'];
$query="select * from login WHERE login_id='".$moha."'";
$res=mysql_query($query);
if(!$res){die(mysql_error());}
while($row=mysql_fetch_array($res)){
$type =$row['type'];
$permission =$row['permission'];
}
if($type==1){
$query="select * from client WHERE login_id='".$moha."'";
$res=mysql_query($query);
if(!$res){die(mysql_error());}
while($row=mysql_fetch_array($res)){
$level_id =$row['level_id'];
$area_id =$row['area_id'];
}
}elseif($type==2){
$query="select * from department WHERE login_id='".$moha."'";
$res=mysql_query($query);
if(!$res){die(mysql_error());}
while($row=mysql_fetch_array($res)){
$level_id =$row['level_id'];
$area_id =$row['area_id'];
}
}elseif($type==3){
$query="select * from clerk WHERE login_id='".$moha."'";
$res=mysql_query($query);
if(!$res){die(mysql_error());}
while($row=mysql_fetch_array($res)){
$level_id =$row['level_id'];
$area_id =$row['area_id'];
}
}
$sql="SELECT * FROM item_price WHERE unit_id='".$q."' && level_id = '".$level_id."' && area_id ='".$area_id."' && item_id='".$gid."' ";
$result = mysql_query($sql);
while($rows = mysql_fetch_array($result)) {
$price = $rows['price'];
echo"<span class='item-price-special'>$price</span>";
}
mysql_close();
?>
【问题讨论】:
-
您所说的“每条记录都超过价值”是什么意思?
-
请发布您在获取 $item_id 时收到的错误消息
-
我的意思是我从数据库中获取它的值,每个项目都有 id ana i 显示超过项目
-
while($row=mysql_fetch_array($res)){ $item_id=$row['item_id'];
-
您说:但是 ID = $item_id 的值没有值,因为 OF 超过了每条记录的值 值是什么样的?
标签: javascript php jquery mysql ajax