【发布时间】:2018-03-27 17:07:20
【问题描述】:
假设我想在页面刷新时对表单中的按钮重新排序,这样当页面刷新时,按钮会重新排列它们的位置。 主要问题是我要在按钮中显示的文本来自数据库数据。 下面是我现在正在尝试的代码-
<?php
session_start();
include_once 'dbconnect.php';
$res=mysqli_query($con, "SELECT * FROM users WHERE user_id=".$_SESSION['user']);
$userRow=mysqli_fetch_assoc($res);
?>
<?php
$array = array(
'<?php echo $userRow['test1']; ?>' => array( 'class' => 'foo', 'name' => 'update' ),
'<?php echo $userRow['test2']; ?>' => array( 'class' => 'john', 'name' => 'update' ),
'<?php echo $userRow['test3']; ?>' => array( 'class' => 'x', 'name' => 'update' ),
'<?php echo $userRow['test4']; ?>' => array( 'class' => 'a', 'name' => 'update' ),
'<?php echo $userRow['test5']; ?>' => array( 'class' => 'b', 'name' => 'update' ),
'<?php echo $userRow['test6']; ?>' => array( 'class' => 'c', 'name' => 'update' ),
'<?php echo $userRow['test7']; ?>' => array( 'class' => 'd', 'name' => 'update' ),
'<?php echo $userRow['test8']; ?>' => array( 'class' => 'e', 'name' => 'update' ),
'<?php echo $userRow['test9']; ?>' => array( 'class' => 'f', 'name' => 'update' ),
'<?php echo $userRow['test10']; ?>' => array( 'class' => 'g', 'name' => 'update' ),
'<?php echo $userRow['test11']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test12']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test13']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test14']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test15']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test16']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test17']; ?>' => array( 'class' => 'h', 'name' => 'idchange' ),
'<?php echo $userRow['test18']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test19']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
'<?php echo $userRow['test20']; ?>' => array( 'class' => 'h', 'name' => 'update' ),
);
uksort( $array, function() { return rand() > rand(); } ); //This will mix them up.
foreach( $array as $item => $atts ){
<form action="check1.php" method="post">
echo '<div class="'. $atts['class'] .'"><button name="'. $atts['name'] .'" type="submit" >'. $item .'</button></div>';
</form>
}
?>
所以,它显示错误.. 那么有谁能帮我解决这个问题吗?
【问题讨论】:
-
作为一个即时错误,我可以看到这里
$array = array( '<?php echo $userRow['test1'];有一个打开的<?php标签,然后打开另一个<?php标签而不关闭第一个标签。请说明您遇到的错误以及用户表中的列。另外请检查您是否提供了Minimal, Complete, and Verifiable example。谢谢。
标签: javascript php html mysql associative-array