使用AJAX，PHP和MySQL显示表数据

Question

我想根据[plan]和[order_id]值显示一列数据[pin]。 计划= 9，订单ID = 0。 想使用ajax加载数据而无需重新加载页面。

这是我的HTML /脚本：

<script>
function showPins(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","getPins.php?q="+str,true);
xmlhttp.send();
}
}
</script>

HTML:
    <div align="center">
    <h3>View PIN's</h3>
    <form>
    <select name="users" onchange="showPins(this.value)">
      <option value="">Select Plan Type:</option>
      <option value="1">Plan1</option>
      <option value="2">Plan2</option>
      <option value="3">Plan3</option>  
    </select>
    </form>
    <br/>
    <div id="txtHint"></div>
    </div>

这是我的PHP文件（getPins.php）：

<?php
$q = intval($_GET['q']);

$con = mysqli_connect('myHost','myUsername','myPw','my_db');
if (!$con) {
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"my_db");
$sql="SELECT * FROM attPins WHERE (order_id=0, plan=9 and id = '".$q."')";
$result = mysqli_query($con,$sql);

echo "<table>
<tr>
<th>PIN's</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
    echo "<tr>";    
    echo "<td>" . $row['pin'] . "</td>";
    echo "</tr>";
}
echo "</table>";
mysqli_close($con);

这基于此处显示的教程： http : //www.w3schools.com/php/php_ajax_database.asp

试图使其针对所选计划类型显示正确的引脚。

Answer 1

您的查询将错误阅读说明书哪里

  $sql="SELECT * FROM attPins WHERE (order_id=0, plan=9 and id = '".$q."')";

这将是

 WHERE (order_id=0 and plan=9 and id = '".$q."')

要么

WHERE (order_id=0 OR plan=9 and id = '".$q."')

根据您的要求