[英]Separate PHP and HTML
我一直在嘗試清理代碼的可讀性,並從此處得到許多答案,盡管我已經走了很遠,但是當我嘗試從PHP拆分HTML時,我的代碼無法正常工作。 當我在PHP代碼塊中使用echo語句時,代碼運行良好。我試圖將存儲的proc的結果輸出到PHP塊外部的HTML表中,但仍使用PHP變量,這是我的代碼:
<?php
include_once ('includes/admin.php');
if (isset($_GET['submit'])) {
$id = $_GET['val1'];
}
$wResult = mysqli_query($con, "call getwisherid($id)");
?>
<html>
<head>
<meta charset="UTF-8">
<title></title>
<link href="sqlcss.css" type="text/css" rel="stylesheet">
</head>
<body>
<form>
<input type="text" name="val1" value="" />
<input type="submit" value="submit" name="submit" />
</form>
<?php while($row = mysqli_fetch_array($wResult)){ ?>
<div class="wish_result">
<table>
<tr>
<td><?php echo $row['name'];?></td>
<td><?php echo $row['password'];?></td>
</tr>
</table>
</div>
</body>
</html>
代碼的一些增強(以及錯誤報告和查詢);-
<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
?>
<html>
<head>
<meta charset="UTF-8">
<title></title>
<link href="sqlcss.css" type="text/css" rel="stylesheet">
</head>
<body>
<form>
<input type="text" name="val1" value="" />
<input type="submit" value="submit" name="submit" />
</form>
<?php
include_once ('includes/admin.php');
if (isset($_GET['val1']) && !empty($_GET['val1'])) {
$id = $_GET['val1'];
if($con){
$wResult = mysqli_query($con, "SELECT name,password FROM <table name> WHERE id = $id"); // put here the appropriate table name
// i don't know what is this :- call getwisherid($id) and it is correct or not
if($wResult){
if(mysqli_num_rows($wResult)>0){
?>
<div class="wish_result">
<table>
<?php while($row = mysqli_fetch_assoc($wResult)){ ?>
<tr>
<td><?php echo $row['name'];?></td>
<td><?php echo $row['password'];?></td>
</tr>
<?php }?>
<?php }else{echo "<tr>No Record Available.</t>";}?>
</table>
<?php }else{"Query error".mysqli_error($con);}?>
</div>
<?php }else{echo "connection error".mysqli_connect_error();}?>
</body>
<?php }else{echo "please fill the form value;"}?>
</html>
您可以將更多內容移入包含文件,或執行以下操作:
<?php
include_once ('includes/admin.php');
if (isset($_GET['submit'])) {
$id = $_GET['val1'];
}
$out = '
<div class="wish_result">
<table>
';
$wResult = mysqli_query($con, "call getwisherid($id)");
while($row = mysqli_fetch_array($wResult)){
$out .= '<tr><td>' .$row['name']. '</td><td>' .$row['password']. '</td></tr>';
}
$out .= '</table></div>';
?>
<html>
<head>
<meta charset="UTF-8">
<title></title>
<link href="sqlcss.css" type="text/css" rel="stylesheet">
</head>
<body>
<form>
<input type="text" name="val1" value="" />
<input type="submit" value="submit" name="submit" />
</form>
<?php echo $out; ?>
</body>
</html>
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.