请有人帮我通过脚本将数据插入mysql数据库时出现解析错误
ive还将其代码改为：
？>
仍然不起作用。。。它显示了与意外结局相同的错误，并提前感谢##
<?php
$conn = connect($config['db']);
if (!$conn)
die("Could not connect to DB");
$user_id = $_SESSION['myuserid'];
$user_row = user_information_row($conn, $user_id);
$routine_names = get_active_routine_names($conn, $user_id);
?>
<style>
.selection_block{
background: rgb(119, 193, 88);
}
</style>
<div class="page">
<div class="information">
<div class="title_nav">
<h1 style="display:inline;">Add Inputs</h1>
<p><a href="input?weights">view</a></p>
<p><a href="input?n">add</a></p>
<p><a href="input?n_edit">edit</a></p>
<p><a href="input">| menu</a></p>
</div>
<form action="" method="post">
<?php
if (count($routine_names)):
?>
<?php
if ($user_row['help']):
?>
<div class="help">
Select the routine you wish to use!
</div>
<?php
endif;
?>
<p>Select today's routine!</p>
<div>
<?php
foreach ($routine_names as $name) {
$exercise_ids_query = $conn->query("SELECT exerciseIdF FROM tbl_routines WHERE userIdF = $user_id AND routineName = '$name' AND active = TRUE");
$exercise_ids = $exercise_ids_query->fetchALL(PDO::FETCH_ASSOC);
$str_exercise_names = "";
foreach ($exercise_ids as $exerciseId) {
$id = $exerciseId['exerciseIdF'];
$exercise_name_query = $conn->query("SELECT name FROM tbl_exercises WHERE exerciseId = $id AND active = true");
$exercise_name = $exercise_name_query->fetchALL(PDO::FETCH_ASSOC);
$str_exercise_names .= $exercise_name[0]['name'] . ', ';
}
$str_exercise_names = substr($str_exercise_names, 0, -2);
$str_exercise_names .= '.';
echo '<a href="input?n&name=' . $name . '"><div class="routine_selection"><p>' . $name . ' - ' . $str_exercise_names . '</p></div></a>';
}
?>
</div>
<?php
else:
?>
<div class="help">
No active routines, click <a href="input.php?r"style="text-decoration: underline; font-weight:bold; color:white">here</a> to make one.
</div>
<?
endif;
?>
</form>
</div>
</div>
<?php
if (isset($_GET['input_status'])){
echo "<div id='"notify'" class='"success'">" . $_GET['input_status'] . "</div>";
}
?>代码的问题是中有一个<?
<?
endif;
?>
如果没有设置短的打开标签，则会导致解析错误。
启用它们，或者将<?更改为<?php
更多关于短开放标签：
- http://php.net/manual/en/language.basic-syntax.phptags.php
- 如何启用PHP短标签
附加说明。
如果您有更多没有向我们展示的代码，并且其中包含<?=，那么这也会增加问题。
您需要将<?=更改为<?php echo，因为<?=是<?php echo的短标记语法。