php - 我如何将每个从 mysql 获取的数据发送到另一个页面
问题描述
我有 2 页“customize.php”和“deepview.php”,
这是 customiz.php:
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.6.4/angular.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<?php
session_start();
?>
<style>
.site {
/* width: 100%; */
height: 100px;
border: 1px solid #ccc;
text-align: center;
padding: 25px;
}
.site:hover {
box-shadow: 0 4px 8px 0 rgba(0, 0, 0, 0.2), 0 6px 20px 0 rgba(0, 0, 0, 0.19);
}
</style>
<div id="vcs"> <!-- view of customize site -->
<?php
include ('connection.php');
$sql = "SELECT site_name FROM sites";
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "<a href='#' id='custlink' data-target='deepview'><div class='col-sm-4 site'>";
echo $row['site_name'];
echo "</div></a>";
$_SESSION['session_site_name'] = $row['site_name'];
}
}
?>
</div>
<script>
$(document).ready(function() {
// custlink is the id of the link
$("a, #custlink").on("click", function() {
var target = $(this).data('target');
$("#vcs").load(target + '.php');
return true;
})
})
</script>
这是 deepview.php:
<?php
session_start();
?>
<style>
.deepsite {
/* width: 100%; */
height: 100px;
border: 1px solid #ccc;
text-align: center;
padding: 25px;
}
.deepsite:hover {
box-shadow: 0 4px 8px 0 rgba(0, 0, 0, 0.2), 0 6px 20px 0 rgba(0, 0, 0, 0.19);
}
</style>
<?php
include ('connection.php');
$clicked_site_name = $_SESSION['session_site_name'];
$sql = "SELECT date from tables WHERE site_name='".$clicked_site_name."'";
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
echo "<a href='#'><div class='col-sm-4 deepsite'>";
echo $row['date'];
echo "</div></a>";
}
}
?>
这里我想要的是当我单击customize.php 上的链接以加载deepview.php 并从mysql 表中选择“日期”,其中它的“site_name”是我在customize.php 中单击的站点
我尝试了上面的代码,它成功加载了 deepview.php,但它总是加载相同的“日期”,即使我在customize.php 中单击了不同的“站点”。
解决方案
在customize.php中,我更改了以前的代码
$sql = "SELECT site_name FROM sites";
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$rr = $row['site_name'];
?>
<a href='#' data-target="database/deepview.php?lo=<?php echo "$rr"; ?>">
<?php
echo "<div class='col-sm-4 site'>";
echo $rr;
echo "</div></a>";
}
}
我在 deepview.php 中使用全局变量 $_GET 提到了“lo”
$sql = "SELECT date from tables WHERE site_name='".$_GET['lo']."'";
$result = mysqli_query($con, $sql);
if(mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
$rr = $row['date'];
?>
<a href='' id='sslink' data-target='database/viewsinglesite?lo=<?php echo "$rr"; ?>'>
<?php
echo "<div class='col-sm-4 deepsite'>";
echo $row['date'];
echo "</div></a>";
}
}
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