【发布时间】：2017-03-01 19:56:35
【问题描述】：
我有一个由数据库中的列填充的下拉列表。在选择时，我希望它显示列等于下拉列表中选择的值的任何表结果。每当我选择一个值时，我都可以显示表头，但没有显示其他数据。
我正在执行echo $q，它返回了正确的值。所以它似乎正在获取选定的值并将其正确传递给test1.php，所以我猜测我的test1.php 脚本中的某处一定有问题。
如何解决此问题，以便显示具有与所选内容相同的 Product Report Code 列的数据库行？
HTML/PHP：
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("txtHint").innerHTML = this.responseText;
}
};
xmlhttp.open("GET","test1.php?q="+str,true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<form>
<section id="rep_dropdown">
<select onchange="showUser(this.value)" name="users">
<option value="">Product Report Code</option>
<?php foreach($repcode->fetchAll() as $reportcode) { ?>
<option value="<?php echo $reportcode['Product Report Code'];?>"><?php echo $reportcode['Product Report Code'];?></option>
<?php } ?>
</select>
</section>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here...</b></div>
</body>
</html>
test1.php：
<html>
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
</head>
<body>
<?php
$q = intval($_GET['q']);
echo $q;
$host="xxxxxxxxx";
$dbName="xxxxxxxxx";
$dbUser="xxxxxxxxxxxxxx";
$dbPass="xxxxxx";
$dbh = new PDO( "sqlsrv:server=".$host."; Database=".$dbName, $dbUser, $dbPass);
$sql="SELECT * FROM vProducts WHERE [Product Report Code] = '".$q."'";
$result = sqlsrv_query($dbh,$sql);
?>
<table id="skuTable" cellspacing="5" class="sortable">
<tr class="ui-widget-header">
<th style="display: none">Product ID</th>
<th class="skuRow">Major Category</th>
<th class="skuRow">Minor Category</th>
<th class="skuRow">Report Code</th>
<th class="skuRow">SKU</th>
<th class="skuRow">SKU Description</th>
<th class="skuRow">SKU Status</th>
<th class="skuRow">Create Date</th>
<th class="skuRow">Group ID</th>
<th class="sorttable_nosort">Edit</th>
</tr>
<?php while($row = sqlsrv_fetch_array($result)) { ?>
<tr class="Row" data-code="<?php echo $row['Product Report Code']?>">
<td style="display: none" class="prod_id" id="product_id-<?php echo intval ($row['Product_ID'])?>"><?php echo $row['Product_ID']?></td>
<td class="major_cat" id="major_cat-<?php echo intval ($row['Major Category'])?>"><?php echo $row['Major Category']?></td>
<td class="minor_cat" id="minor_cat-<?php echo intval ($row['Minor Category'])?>"><?php echo $row['Minor Category']?></td>
<td class="rep_code" id="rep_code-<?php echo intval ($row['Product Report Code'])?>" align="center"><?php echo $row['Product Report Code']?></td>
<td class="sku" id="sku-<?php echo intval ($row['SKU'])?>" align="center"><?php echo $row['SKU']?></td>
<td class="sku_desc" id="sku_desc-<?php echo intval ($row['SKU Description'])?>"><?php echo $row['SKU Description']?></td>
<td class="sku_status" id="sku_status-<?php echo intval ($row['SKU Status'])?>" align="center"><?php echo $row['SKU Status']?></td>
<td class="create_date" id="create_date-<?php echo intval ($row['Date'])?>" align="center"><?php echo $row['Date']?></td>
<td class="group_id" id="group_id-<?php echo intval ($row['Group_ID'])?>" align="center"><?php echo $row['Group_ID']?></td>
<td><input type="button" class="edit" name="edit" value="Edit" onclick="enable_value(this)"></td>
</tr>
<?php } ?>
</table>
</body>
</html>
【问题讨论】：
标签： php jquery html sql-server