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我开发了一个 PHP 页面，显示 MySQL 表中的所有项目。现在我想在用户单击此页面上的某个项目时显示一个特定项目。如何获取此特定 ID 以及如何配置<a href="">链接？

更新问题：

这是 cars.php 页面的标题（包含所有结果）：

<?

require("admin/db/connect.php");

$sql = "SELECT * FROM tb_carros";

$limite = mysql_query("$sql");

$dados = array();
while ($sql = mysql_fetch_array($limite) ) {
    $dados[] = $sql;
}

?>

和 HTML：

    <?php foreach ($dados as $row): ?>

    <div id="containerResumo">

        <a href="#"> <!-- this is the key problem -->

        <div class="dadosResumo">
            <?=$row['carro']?><br /><br />
            Ano:   <?=$row['ano']?><br /><br />
            Câmbio: <?=$row['cambio']?><br /><br />
            R$ <?=$row['valor']?>
        </div><!-- END of dadosItem -->
        </a>

    </div><!-- END of containerResumo -->

    <?php endforeach ?>

现在，当用户单击一个项目时，我想打开页面 carro_item.php 并在其上加载项目的数据。

数据库中的参考 ID 是 id_carro

我尝试了许多类型的代码，但没有奏效。即使我将完整的网址放在浏览器上，也不会加载数据：

http://adrianomachado.com/testesClientes/carro_item.php?id_carro=1

这是 carro_item.php 的 PHP：

<?

require("admin/db/connect.php");

$id = (int)$_GET['id_carro'];

$sql = "SELECT * FROM tb_carros WHERE id = $id";

?>

和 HTML：

            <div class="dadosItem">
                R$ <?php $valor ?><br /><br />
                Ano:   <?php $ano ?><br /><br />
                Kilometragem: <?php $km ?><br /><br />
                Cor: <?php $cor ?><br /><br />
                Portas: <?php $portas ?><br /><br />
                Combustível: <?php $combustivel ?><br /><br />
                Câmbio: <?php $cambio ?><br /><br />
                Final da placa: <?php $final_placa ?><br /><br />
                Carroceria: <?php $carroceria ?>
            </div><!-- END of dadosItem -->

有什么帮助吗？

更新 02：

这是 carro_item 中的查询：

<?

require("admin/db/connect.php");

$sql = "SELECT * FROM tb_carros";

$limite = mysql_query("$sql");

$dados = array();
while ($sql = mysql_fetch_array($limite) ) {
    $dados[] = $sql;
}


?>

但是，很明显它会返回所有结果，例如 cars.php 页面。问题是如何将结果过滤到用户单击的链接的相同 ID？

我不知道如何编写 $sql = "SELECT * FROM tb_carros"; 行来做到这一点。

4

2 回答 2

1

您可以使用$_GETvariables。

如果页面上的链接格式如下：

<a href="mypage.php?id=5">

然后在您的 php 页面中，您将能够通过全局$_GET数组访问该值。

$id = mysqli_real_escape_string($_GET['id']); // $id will have the value passed to it by the link

请注意不要让自己对SQL 注入开放，尽管通过清理参数或使用参数化查询。

参考

编辑：

To create the links formatted in the correct way, you'd first retrieve all of the ids you need and store them in an array. I'll use $ids as an example.

$ids = array(1, 50, 25, 62, ...); // This was populated from the database

// Loop through all ids and output link code for each one
foreach ($ids as $link_id) {
  echo '<a href="mypage.php?id=' . $link_id . '">Click me</a>';
}

Edit2:

In cars.php format the links like this:

<a href="/testesClientes/carro_item.php?id_carro=<?= $row['id'] ?>">

Edit3:

Your carro_item.php should look something like this:

<?php

  require("admin/db/connect.php");

  $id = (int)$_GET['id_carro'];

  $sql = "SELECT * FROM tb_carros WHERE id = $id";

  $result = mysql_query($sql);

  $row = mysql_fetch_array($result);

  // ...
?>

<!-- And your HTML should look something like this -->
<!-- ... -->

<div class="dadosItem">
  R$ <?= $valor ?><br /><br />
  Ano:   <?= $row['ano'] ?><br /><br />
  Kilometragem: <?= $row['km'] ?><br /><br />
  Cor: <?= $row['cor'] ?><br /><br />
  Portas: <?= $row['portas'] ?><br /><br />
  Combustível: <?= $row['combustivel'] ?><br /><br />
  Câmbio: <?= $row['cambio'] ?><br /><br />
  Final da placa: <?= $row['final_placa'] ?><br /><br />
  Carroceria: <?= $row['carroceria'] ?>
</div><!-- END of dadosItem -->

<!-- ... -->

Also, you should stray from using functions of the form mysql_* as they are deprecated. See Why shouldn't I use mysql_* functions in PHP? for more information.

于 2013-04-07T21:44:06.347 回答
0

PHP：

 $id = (int)$_GET['id'];

HTML：

<a href="page.php?id=<?= $id; ?>">

询问：

SELECT * FROM table WHERE id = $id;
于 2013-04-07T21:44:05.280 回答