javascript - 通过 php 查询链接到弹出 View

Question

您好，我的 index.php 页面中有一个搜索功能。当用户在字段中输入“搜索”时，它会转到 jobsearch.php 并从 mysqli 数据库中打印出工作摘要。

我已经创建了结果末尾的链接。 因此，当用户看到职位摘要时，用户会点击链接(点击此处)，然后将用户定向到一个弹出 View ，其中显示有关该职位的更多信息(职位_描述)。是否有这样做的方式？我不想为每个职位创建 html 页面，因为数据库中可能有 100 个职位。

我希望点击此处链接获取所需职位的 ID 并在弹出页面上打印出职位描述。

看起来像这样:

这是打印作业列表的 jobs.php..

<?php
$servername = "*****";
$username = "root";
$password = "*****";
$dbname = "jobslist";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, job_title, job_description, job_location, job_category  FROM jobs_list";
$result = $conn->query($sql);

//display table
echo "<table border='1'>
<!--<tr>
<th>ID</th>
<th>Title</th>
<th>Description</th>
<th>Location</th>
<th>Category</th>
</tr>-->";

if ($result->num_rows > 0) {
     // output data of each row

     while($row = $result->fetch_assoc()) {

 echo "<tr>";
       echo "<td min-height:'200' ><h2>". $row["id"] ."</h2></td>";
       echo "<td>". $row["job_title"] . "</td>";
       echo "<td min-width:'700' >". $row["job_description"] . "</td> " ;
       echo "<td>". $row["job_location"] . "</td>";
       echo "<td>". $row["job_category"] . "</td> " ;

//this prints out a clickable link for every job

echo "<td><a href='" . $row['id'] .  "'>Click Here</a></td>";



       echo "</tr>";
    }

}
 else {
     echo "0 results";
}

echo "</table>";
$conn->close();
?>

Answer 1

改变

//this prints out a clickable link for every job

echo "<td><a href='" . $row['id'] .  "'>Click Here</a></td>";

到

echo "<td><a target='jobdescriptionwindow' href='jobdescription.php?id=" . $row['id'] .  "'>Click Here</a></td>";

当您点击将所点击职位的 ID 传递给 jobdescription.php 的链接时，这将打开一个新窗口(称为“jobdescriptionwindow”)。您还必须确保您的浏览器不会阻止此窗口打开。

使用以下内容创建一个名为 jobdescription.php 的新 PHP 文档:

<?php

if (isset($_GET['id'])) { // Check ID is is present in parameters

    $servername = "*****"; // <-- Enter your details
    $username = "root";
    $password = "*****"; // <-- Enter your details
    $dbname = "jobslist";

    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);

    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }

    // I do: (int)$_GET['id'] to type cast the value of $_GET['id'] to protect against injection. http://php.net/manual/en/language.types.type-juggling.php

    $sql = "SELECT id, job_title, job_description, job_location, job_category FROM jobs_list WHERE id = ".(int)$_GET['id'];

    $result = $conn->query($sql); // <-- Added new line
    $row = $result->fetch_assoc();


    echo $row['job_title'] . '<br />';
    echo $row['job_description']; //<-- Does now show your description?

    var_dump($result); // Lay out this data as required
    // Outputs object(mysqli_result)#2 (5) { ["current_field"]=> int(0) ["field_count"]=> int(5) ["lengths"]=> array(5) { [0]=> int(1) [1]=> int(23) [2]=> int(102) [3]=> int(4) [4]=> int(2) } ["num_rows"]=> int(1) ["type"]=> int(0) } 



} else {

  echo "No ID provided";

}

这将获取在 GET 参数中发送的作业 ID 的详细信息。