Question

我在mysql数据库中有不同的表，如表1到表5。当我从下拉表列表中选择它时，我想要一个显示每个表的代码。当我尝试执行以下时，我没有得到任何响应，也没有从sql加载数据。

<div> 
<form action="table1.php" method="GET">
<input list="name" name="name">
<datalist id="name">
    <option value="table 1">
    <option value="table 2">
    <option value="table 3">
    <option value="table 4">
    <option value="table 5">
</datalist>
<input type="submit" name="search" value="search">
</form>
</div>

<div>
<table width="600" border="0" cellpadding="1" cellspacing="1">
    <tr>
        <th>Column 1</th>
        <th>Column 2</th>
        <th>Column 3</th>
        <th>Column 4</th>
    </tr>
</table>

在php代码中我为选择值分配一个变量并在程序中给出。但我得到一个错误，该变量未定义或不在$ conn语句中使用。

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sheet";
$tbname= $_GET['name'];

if (isset($_GET['search'])) {
    try {
        $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
        $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $stmt = $conn->prepare("SELECT * FROM '$tbname'");
        $stmt->execute();

        $result = $stmt->setFetchMode(PDO::FETCH_ASSOC);

        echo "<tr>";
        echo "<th>".$result['Column 1']."</th>";
        echo "<th>".$result['Column 2']."</th>";
        echo "<th>".$result['Column 3']."</th>";
        echo "<th>".$result['Column 4']."</th>";
        echo "</tr>";
    }
    catch(PDOException $e) {
        echo "Error: " . $e->getMessage();
    }
    $conn = null;
    echo "</table>";
}

Answer 1

您已将表名放在单引号中，这是不正确的，您需要使用反引号。

PHP代码中的以下行是错误的

$stmt = $conn->prepare("SELECT * FROM '$tbname'");

应该改为

$stmt = $conn->prepare("SELECT * FROM `$tbname`");

Answer 2

您在此处所做的一切都设置了获取模式，您实际上并没有获取任何内容。

变化：

$stmt->setFetchMode(PDO::FETCH_ASSOC);

echo "<tr>";
echo "<th>".$result['Column 1']."</th>";
echo "<th>".$result['Column 2']."</th>";
echo "<th>".$result['Column 3']."</th>";
echo "<th>".$result['Column 4']."</th>";
echo "</tr>";

为：

while ($result = $stmt->fetch(PDO::FETCH_ASSOC)) {
     echo "<tr>";
     echo "<th>".$result['Column 1']."</th>";
     echo "<th>".$result['Column 2']."</th>";
     echo "<th>".$result['Column 3']."</th>";
     echo "<th>".$result['Column 4']."</th>";
     echo "</tr>";
}

Answer 3

希望这段代码可以帮到你，

我稍微改变了你的代码结构并添加了fetchAll()方法来从db中检索行，你也可以使用fetch()方法逐行获取最后我打印这些行最后。

<?php 
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "sheet";
$tbname= $_GET['name'];
$results = array();

if (isset($_GET['search'])) {
    try {
        $conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
        $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $stmt = $conn->prepare("SELECT * FROM `$tbname`");
        $stmt->execute();

        $stmt->setFetchMode(PDO::FETCH_ASSOC);
        $results = $stmt->fetchAll();//Add this line to get db results

    } catch(PDOException $e) {
        echo "Error: " . $e->getMessage();
    }
    $conn = null;
}
?>

<div> 
    <form action="table1.php" method="GET">
        <input list="name" name="name">
        <datalist id="name">
            <option value="table 1">
            <option value="table 2">
            <option value="table 3">
        </datalist>
        <input type="submit" name="search" value="search">
    </form>
</div>

<div>
    <table width="600" border="0" cellpadding="1" cellspacing="1">
        <tr>
            <th>Column 1</th>
            <th>Column 2</th>
            <th>Column 3</th>
            <th>Column 4</th>
        </tr>
        <?php
        if (is_array($results) && count($results)>0) {
            foreach ($results as $result) {
                echo "<tr>";
                echo "<td>".$result['Column 1']."</td>";
                echo "<td>".$result['Column 2']."</td>";
                echo "<td>".$result['Column 3']."</td>";
                echo "<td>".$result['Column 4']."</td>";
                echo "</tr>";
            }
        }
        ?>
    </table>
</div>

如何使用php中的下拉列表显示不同的mysql表

3 个答案: