PHP 语法错误“意外的 $end”
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PHP syntax error “unexpected $end”
提问by Hymansta
I have 3 files 1) show_createtable.html 2) do_showfielddef.php 3) do_showtble.php
我有 3 个文件 1) show_createtable.html 2) do_showfielddef.php 3) do_showtble.php
1) First file is for creating a new table for a data base, it is a fom with 2 inputs, Table Name and Number of Fields. THIS WORKS FINE!
1) 第一个文件用于为数据库创建一个新表，它是一个具有 2 个输入、表名和字段数的表单。这很好用！
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<h1>Step 1: Name and Number</h1>
<form method="post" action="do_showfielddef.php" />
<p><strong>Table Name:</strong><br />
<input type="text" name="table_name" size="30" /></p>
<p><strong>Number of fields:</strong><br />
<input type="text" name="num_fields" size="30" /></p>
<p><input type="submit" name="submit" value="go to step2" /></p>
</form>
</body>
</html>
2) this script validates fields and createa another form to enter all the table rows. This for also WORKS FINE!
2）此脚本验证字段并创建另一个表单以输入所有表格行。这也很好用！
<?php
//validate important input
if ((!$_POST[table_name]) || (!$_POST[num_fields])) {
header( "location: show_createtable.html");
exit;
}
//begin creating form for display
$form_block = "
<form action=\"do_createtable.php\" method=\"post\">
<input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\">
<table cellspacing=\"5\" cellpadding=\"5\">
<tr>
<th>Field Name</th><th>Field Type</th><th>Table Length</th>
</tr>";
//count from 0 until you reach the number fo fields
for ($i = 0; $i <$_POST[num_fields]; $i++) {
$form_block .="
<tr>
<td align=center><input type=\"texr\" name=\"field name[]\"
size=\"30\"></td>
<td align=center>
<select name=\"field_type[]\">
<option value=\"char\">char</option>
<option value=\"date\">date</option>
<option value=\"float\">float</option>
<option value=\"int\">int</option>
<option value=\"text\">text</option>
<option value=\"varchar\">varchar</option>
</select>
</td>
<td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\">
</td>
</tr>";
}
//finish up the form
$form_block .= "
<tr>
<td align=center colspan=3><input type =\"submit\" value=\"create table\">
</td>
</tr>
</table>
</form>";
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create a database table: Step 2</title>
</head>
<body>
<h1>defnie fields for <? echo "$_POST[table_name]"; ?>
</h1>
<? echo "$form_block"; ?>
</body>
</html>
Problem is here 3) this form creates the tables and enteres them into the database. I am getting an error on line 37 "Parse error: syntax error, unexpected $end in /home/admin/domains/domaina.com.au/public_html/do_createtable.php on line 37"
问题在这里 3) 这个表格创建表格并将它们输入到数据库中。我在第 37 行收到错误“解析错误：语法错误，第 37 行在 /home/admin/domains/domaina.com.au/public_html/do_createtable.php 中意外 $end”
<?
$db_name = "testDB";
$connection = @mysql_connect("localhost", "admin_user", "pass")
or die(mysql_error());
$db = @mysql_select_db($db_name, $connection)
or die(mysql_error());
$sql = "CREATE TABLE $_POST[table_name](";
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
if ($_POST[field_length][$i] !="") {
$sql .=" (".$_POST[field_length][$i]."),";
} else {
$sql .=",";
}
$sql = substr($sql, 0, -1);
$sql .= ")";
$result = mysql_query($sql, $connection) or die(mysql_error());
if ($result) {
$msg = "<p>" .$_POST[table_name]." has been created!</p>";
?>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create A Database Table: Step 3</title>
</head>
<body>
<h1>Adding table to <? echo "$db_name"; ?>...</h1>
<? echo "$msg"; ?>
</body>
</html>
回答by x4tje
$result = mysql_query($sql, $connection) or die(mysql_error());
if ($result) {
$msg = "<p>" .$_POST[table_name]." has been created!</p>";
}
you missing a }in your last if statement, and your for loop is missing a }too
你错过了}在过去的if语句，和你的循环缺少}过
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
if ($_POST[field_length][$i] !="") {
$sql .=" (".$_POST[field_length][$i]."),";
} else {
$sql .=",";
}
}
回答by Gumbo
This error message means that a control structure block isn't closed properly. In your case the closing }of some of your control structures like the forloop or the last ifare missing.
此错误消息意味着控制结构块未正确关闭。在您的情况下，缺少}某些控制结构（例如for循环或最后一个）的关闭if。
You should use proper indentation and an editor that highlights bracket pairs to have a visual aid to avoid such errors.
您应该使用适当的缩进和突出显示括号对的编辑器，以获得避免此类错误的视觉帮助。
回答by yassin
You must close the for expression block:
您必须关闭 for 表达式块：
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
// ...
}
回答by user3095614
?
in your php.ini(php configuration) change :
? 在您的php.ini（php 配置）更改中：
short_open_tag = Off
short_open_tag = 关闭
you opened php tag shortly at line 1
just find and replace all <?with <?php
您在第 1 行很快打开了 php 标签，
只需找到并替换所有<? 用<?php
回答by codaddict
Your forloop is not terminated. You are missing a }
你的for循环没有终止。你缺少一个}
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
}
for ($i = 0; $i < count($_POST[field_name]); $i++) {
$sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];
}
And as pointed by others there is also a missing }for the last ifstatement:
正如其他人所指出的}，最后一条if语句也有缺失：
if ($result) {
$msg = "< p>" .$_POST[table_name]." has been created!< /p>";
}
if ($result) {
$msg = "< p>" .$_POST[table_name]." has been created!< /p>";
}
回答by Marc B
Stylistic tip: Use HEREDOCs to assign blocks of text to a variable, instead of the hideous multi-line-with-tons-of-escaping-backslashes constructs you're using. They're far easier to read and less error prone if/when you happen to forget a \ somewhere and break the script with a parse error.
风格提示：使用HEREDOC将文本块分配给变量，而不是您正在使用的可怕的多行转义反斜杠构造。如果/当您碰巧在某处忘记了 \ 并使用解析错误破坏脚本时，它们更容易阅读且不易出错。
回答by Anil
I just solved this error, after checking my code, I had no open tags/braces.
For me, I got this error when moving to a amazon server.
It turns out I needed to enable short_open_tag = Onin my php.ini.
This solved this error for me.
我刚刚解决了这个错误，在检查我的代码后，我没有打开的标签/大括号。
对我来说，我在移动到亚马逊服务器时遇到了这个错误。
事实证明我需要short_open_tag = On在我的php.ini.
这为我解决了这个错误。
