Question

我正在尝试显示数据onselect方法但数据未显示。这是代码。

的index.html

<html>
<head>
<script>
function showUser(str)
{
    alert(str);
    if (str=="")
    {
        document.getElementById("txtHint").innerHTML="";
        return;
    } 
    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","getuser.php?q="+str,true);
    xmlhttp.send();
}
</script>
</head>
<body>

<form>
    <select name="users" onchange="showUser(this.value)">
        <option value="">Select a person:</option>
        <option value="1">MEGHALI</option>
        <option value="2">SONY</option>
        <option value="3">DHANU</option>
        <option value="4">ADITYA</option>
        <option value="5">MAHESH</option>
        <option value="6">NEERAJ</option>
        <option value="7">VINEET</option>
        <option value="8">ASHWINI</option>
        <option value="9">SANTOSH</option>
    </select>
</form>
<br>
<div id="txtHint"><b>Person info will be listed here.</b></div>

</body>
</html>

和getuser.php

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

$con = mysqli_connect("localhost", "root", " ", "flowers");
if (!$con)
{
    die('Could not connect: ' . mysqli_error($con));
}

mysqli_select_db($con,"FLOWERS");
$sql="SELECT * FROM STUDENTS WHERE id = '".$q."'";

$result = mysqli_query($con,$sql);

echo "<table border='1'>
<tr>
<th>ID</th>
<th>NAME</th>
<th>CITY</th>
<th>PHONE</th>

</tr>";

while($row = mysqli_fetch_array($result))
{
    echo "<tr>";
    echo "<td>" . $row['ID'] . "</td>";
    echo "<td>" . $row['NAME'] . "</td>";
    echo "<td>" . $row['CITY'] . "</td>";
    echo "<td>" . $row['PHONE'] . "</td>";

    echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

我猜getuser.php?q="+str没有通过。
请帮忙，因为我是ajax的新手。

Answer 1

如果你在getuser.php中传递id，如：

getuser.php Q = 1

你应该传递参数，如：

$q = $_GET['q'];

并查询数据库

$sql="SELECT * FROM STUDENTS WHERE id = '".$q."'";

Answer 2

简单地传递不会得到你需要的变量

$q=$_GET['q'];

在您的查询页面中。然后再试一次。如果仍然无法正常工作，请尝试回显您的mysql查询并检查是否在那里获得$ q的值。

无法显示mysql中的数据

2 个答案: