php/mysql选择其中id=url id
我的脚本有问题。我有一个项目页面,如果您单击其中一个项目(这是用户的项目列表),您将使用project.php?project=id进入项目页面 项目scipt:php/mysql选择其中id=url id,php,mysql,Php,Mysql,我的脚本有问题。我有一个项目页面,如果您单击其中一个项目(这是用户的项目列表),您将使用project.php?project=id进入项目页面 项目scipt: <?php require_once 'core/init.php'; include("header.php"); ?> <?php $project = DB
<?php
require_once 'core/init.php';
include("header.php");
?>
<?php
$project = DB::getInstance()->query("SELECT * FROM projects WHERE user_id = '" . $_SESSION['user'] . "' ");
$project_item = new Projectitem();
if(!$project->count()) {
echo 'No Projects';
} else {
?>
<div class="table-responsive">
<table class="table table-bordered">
<?php
foreach($project->results() as $project) {
echo "<tr>";?>
<td width="9,375em"><img src="<?php echo $project->project_img; ?>"><?php
echo "</td>";
echo "<td>";?>
<a href="project_item.php?project=<?php echo escape($project->id); ?>"><?php echo escape($project->name); ?></a>
<?php
echo "</td>";
echo"</tr>";
}
}
echo "</table>";
echo "</div>";
include("footer.php");
?>
您需要将参数绑定到查询,然后执行查询。如果你按照计划去做,你很快就能解决这个问题。您正在
$\u GET['project']
中查找id,因为您正在通过URL的查询字符串传递变量。请您帮助我处理代码,我需要做些什么:很难解析转储到注释中的代码。编辑您的原始帖子并在那里添加更新的信息。
<?php
require_once 'core/init.php';
include("header.php");
?>
<?php
$project = DB::getInstance()->query("SELECT * FROM projects WHERE id = ? ");
if(!$project->count()) {
echo 'No Projects';
} else {
?>
<div class="table-responsive">
<table class="table table-bordered">
<?php
foreach($project->results() as $project) {
echo "<tr>";?>
<?php echo "<td>";?>
<?php echo escape($project->name); ?>
<?php
echo "</td>";
echo"</tr>";
}
}
echo "</table>";
echo "</div>";
include("footer.php");
?>
SELECT * FROM projects WHERE id = ? //(the id from the url)