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Question

我在PHP / HTML / SQLITE3中工作。我有一个由几个表组成的数据库，其中一个表叫做Item，它包含itemID，item的名称等等。所以我的搜索接受了itemID的用户输入，我想要返回的是与该itemID相关的所有内容。

我已经测试了我的搜索，它确实返回了itemID，但是，我在查找如何返回与itemID相关的所有内容时遇到了一些麻烦。下面是我的搜索表单以及我对包含查询的单独文件的内容。

<form method="POST" action="action.php">
    <input name="search" type="text" size="20" maxlength="10"/>
    <input type="submit" value="Search"/>
</form>

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<?php
    if (isset($_POST["search"])) {
        $itemID = $_POST["search"];

    try {
        $db->beginTransaction();
        $query = "SELECT * FROM Item WHERE itemID = '$itemID';";
        $result = $db->query($query);

        if (empty($_POST['search'])){
            echo "<strong>You didn't fill in anything!</strong>";
        }
        else {
            echo $itemID;
        }

        $db->commit();
        }
    $db = null;
    ?>

编辑代码（添加对fetchall的尝试）：

<?php
if (isset($_POST["search"])) {
    $itemID = $_POST["search"];

try {
    $db->beginTransaction();
    $query = "SELECT * FROM Item WHERE itemID = '$itemID';";
    #$result = $db->query($query);
    $result = sqlite_fetch_all($query, SQLITE_ASSOC);

    foreach($result as $entry) {
        echo 'ItemID: ' . $entry['itemID'] . ' Item Name' . $entry['name'];
    }

    if (empty($_POST['search'])){
        echo "<strong>Esteemed customer did not fill in a 
        itemID number, please search again. </strong>";
    }

    $db->commit();
    }

第二次尝试：

 <?php
    $dbname = "mydatabase.db";

  try {
    // Establish connection to "mydatabase.db"
    $db = new PDO("sqlite:" . $dbname);

    // Set error handling so that errors throw an exception
    $db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

    // Enable foreign key constraints on the database
    $db->exec("PRAGMA foreign_keys = ON;");
  } catch (PDOException $e) {
    echo "SQLite connection failed: " . $e->getMessage();
    exit();
  }

    if (isset($_POST["search"])) {
        $itemID = $_POST["search"];

    try {
        $sth = $db->prepare("SELECT * FROM Item WHERE itemID = '$itemID'");
        #$query = "SELECT * FROM Item WHERE itemID = '$itemID';";
        #$result = $db->query($query);
        $sth->execute();
        $result = $sth->fetchAll();
        print_r($result);

        #if (empty($_POST['search'])){
            #echo "<strong>Esteemed customer did not fill in a 
            #itemID number, please search again. </strong>";
        }
    }

    ?>

非常感谢任何输入。

Answer 1

您应该将itemid连接到查询

$query = "SELECT * FROM Item WHERE itemID = '" . $itemID . "';";

PHP - 使用表单和返回记录进行搜索

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1 个答案: