Question

我选择选择值时尝试从数据库中提取值。此选择也从同一数据库填充。

要明确的是，我对AJAX请求不熟悉，并且不太了解我做错了什么。当值出现时，我得到的只是空白。

这是我的html表单：

＆＃13;

function showStock(str) {
    if (str == "") {
        document.getElementById("stockValue").innerHTML = "";
        return;
    } else {
        xmlhttp = new XMLHttpRequest();
        xmlhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
                document.getElementById("stockValue").innerHTML = this.responseText;
            }
        };
        xmlhttp.open("GET", "getstock.php?q=" + str, true);
        xmlhttp.send();
    }
}

＆＃13;

<!DOCTYPE html>
<html>

<head>
</head>

<body>

    <?php $link=mysqli_connect( "localhost", "root", "", "cocktails"); // Check connection if($link===f alse){ die( "ERROR: Could not connect. " . mysqli_connect_error()); } ?>

    <script src='menu.js'></script>

    <form action="addstock.php" method="post">
        Product:
        <br>
        <select name="products" onChange="showStock(this.value)">
            <option value="">Select a product:</option>
            <?php $sql=m ysqli_query($link, "SELECT name FROM inventory ORDER BY name"); while ($row=$ sql->fetch_assoc()){ echo "
            <option value=\ "".$row[ 'name']. "\">" . $row['name'] . "</option>"; } ?>
        </select>
        <br> Stock Value:
        <div id="stockValue">
            <b>0</b>
        </div>

    </form>
</body>

</html>

＆＃13;

这是我的php文件：

<!DOCTYPE html>
<html>

<head> </head>

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

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

        $sql="SELECT * FROM inventory WHERE name = '".$q."'";
        $result = mysqli_query($con,$sql);
        $row = mysqli_fetch_array($result);

        echo $row['stock'];
        mysqli_close($con);
        ?>
</body>

</html>

Answer 1

嗯，这不是那么明显但你的问题是：

$q = intval($_GET['q']);

您的$_GET['q']是一个字符串，您将其转换为intval，因此它变为0。

由于这个原因，你的SQL查询一无所获。

将此行更改为$q = $_GET['q'];，它将起作用：）

Answer 2

从你的php中删除所有HTML，我认为一切都会好的。你的php响应必须是plainText，而不是HTML

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

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

    $sql="SELECT * FROM inventory WHERE name = '".$q."'";
    $result = mysqli_query($con,$sql);
    $row = mysqli_fetch_array($result);

    echo $row['stock'];
    mysqli_close($con);
    ?>

Answer 3

<?php $link=m ysqli_connect( "localhost", "root", "", "cocktails"); // Check connection if($link===f alse){ die( "ERROR: Could not connect. " . mysqli_connect_error()); } ?>

应该是..

<?php $link=mysqli_connect( "localhost", "root", "", "cocktails"); // Check connection if($link===f alse){ die( "ERROR: Could not connect. " . mysqli_connect_error()); } ?>

可能是一个错字，但mysqli_connect上的空格将导致ti失败

再次在这里..

<?php $sql=m ysqli_query($link, "SELECT name FROM inventory ORDER BY name");

应该是..

<?php $sql=mysqli_query($link, "SELECT name FROM inventory ORDER BY name");

来自select表单的Ajax请求

3 个答案: