Question

我有Yii2应用程序，我的数据库中有这样的表：

我试图通过jquery执行sqlQuery。我在 index.php ：

中设置了view和jQuery

<html>
<head>
    <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", "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">Alfredo Jangkirik</option>
            <option value="2">Budi Wunduk</option>
        </select>
    </form>
    <br>
    <div id="txtHint"><b>Person info will be listed here...</b></div>

</body>

并在 getuser.php 中设置查询和数据库连接：

<!DOCTYPE html>
<html>
<head>
    <style>
        table {
            width: 100%;
            border-collapse: collapse;
        }

        table, td, th {
            border: 1px solid black;
            padding: 5px;
        }

        th {text-align: left;}
    </style>
</head>
<body>

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

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

    mysqli_select_db($con, "ajax_demo");
    $sql = "SELECT * FROM user WHERE id = '" . $q . "'";
    $result = mysqli_query($con, $sql);

    echo "<table>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Hometown</th>
<th>Job</th>
</tr>";
    while ($row = mysqli_fetch_array($result)) {
        echo "<tr>";
        echo "<td>" . $row['FirstName'] . "</td>";
        echo "<td>" . $row['LastName'] . "</td>";
        echo "<td>" . $row['Age'] . "</td>";
        echo "<td>" . $row['Hometown'] . "</td>";
        echo "<td>" . $row['Job'] . "</td>";
        echo "</tr>";
    }
    echo "</table>";
    mysqli_close($con);
    ?>
</body>

但它没有工作，如何使其工作以及为什么它不起作用？ 我需要通过jQuery从Sql获取价值。我在原生项目中尝试过这种方法（不使用yii2框架）并且运行良好。

由于

通过jQuery进行sql查询并没有在Yii2中工作，但在本机中工作得很好

0 个答案: