php - string(49) "select * from php mysql 错误

Question

我正在制作一个网页，该网页旨在显示数据库中的数据。数据库存储在 MySQL 中，我正在用 PHP 制作网页。我的 PHP 代码是

<form action="list_projects.php" method="post">
    <p>Choose Search Type: <br /></p>
    <select name="searchtype">
        <option value="partNo">Part Number</option>
        <option value="pname">Part Name</option>
        <option value="color">Part Colour</option>
        <option value="weight">Part Weight</option>
        <option value="city">City</option>
    </select>
    <br />
    <p>Enter Search Term: </p>
    <br />
    <input name="searchterm" type="text" size="20"/>
    <br />
    <input type="submit" name="submit" value="Search"/>
</form>

<?php
    $searchtype=$_POST['searchtype'];
    $searchterm=trim($_POST['searchterm']);
    if (!$searchtype || !$searchterm) {
        echo 'No search details. Go back and try again.';
        exit;
    }

    $query = "select * from project where ".$searchtype." like '%".$searchterm."%'";
    var_dump($query);

    $result = mysqli_query($link,$query);
    $num_results = mysqli_num_rows($result);

    echo "<p>Number of projects found: ".$num_results."</p>";
    for ($i=0; $i <$num_results; $i++) {
        $row = mysqli_fetch_assoc($result);
        echo "<p><strong>".($i+1).". Part Number: ";
        echo htmlspecialchars(stripslashes($row['partNo']));
        echo "</strong><br />Part Name: ";
        echo stripslashes($row['pname']);
        echo "<br />Part Colour: ";
        echo stripslashes($row['color']);
        echo "<br />Part Weight: ";
        echo stripslashes($row['weight']);
        echo "<br />City";
        echo stripcslashes($row['city']);
        echo "</p>";
    }
    mysqli_free_result($result);
    mysqli_close($link);
?>

但是当我运行它时，我得到 string(49) "select * from project where projectNo like '%J1%'"Number of projects found: This PHP脚本旨在加载数据库中的不同项目，并在调用此脚本的 welcome.php 脚本中加载连接到数据库并且它确实正确连接到数据库。

Answer 1

看起来您 var dumped 了错误的变量。你可以试试这个:

$query = "SELECT * FROM project WHERE ".$searchtype." LIKE '%".$searchterm."%'";

$result = mysqli_query($link,$query) or die("Line ".__LINE__." Error found: ".mysqli_error($link)); // If there's an error, it should show here.