PHP错误：“语法错误，意外的T_STRING，期望为'，'或';'”
[英]PHP error: “syntax error, unexpected T_STRING, expecting ',' or ';'”
问题描述
I got this message (syntax error, unexpected T_STRING, expecting ',' or ';') when I was trying to run the following code: 
当我尝试运行以下代码时，收到此消息（语法错误，意外的T_STRING，期望为“;”或“;”）：
<html>
<head>
</head>
<body>
<title>Num One Website</title>
<?
$con = mysql_connect("","","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("", $con);
$result1 = mysql_query("SELECT * FROM Students") ;
echo "<form action='ConfirmEnter.php' method='post'>";
echo "<input type="Radio" name="mark" value="mt">"."MidTerm<br>";
echo "<input type="Radio" name="mark" value="pr">"."Project<br>";
echo "<input type="Radio" name="mark" value="fi">"."Final<br>";
echo "<table border cellpadding=3>";
echo "<tr>";
echo "<th>ID</th>";
echo "<th>MidTerm</th>";
echo "<th>Project</th>";
echo "<th>Final</th>";
echo "<th>Total</th>". "</tr>";
$count=1;
while($row1 = mysql_fetch_array($result1))
{
echo "<tr>";
echo "<td><input name='ID[]' readonly='readonly' value='". $row1['ID'] ."' size=5/></td> ";
echo "<td><input type='text' name='mt[]' size=5 value='0.0' /></td>";
echo "<td><input type='text' name='pr[]' size=5 value='0.0' /></td>";
echo "<td><input type='text' name='fi[]' size=5 value='0.0' /></td>";
echo "</tr>";
$count++;
}
echo "</table>";
echo "<input type='submit' value='Submit' />";
echo "</form>";
mysql_close($con);
?>
</body>
</html>
The error in these lines: 这些行中的错误：
echo "<input type="Radio" name="mark" value="mt">"."MidTerm<br>";
echo "<input type="Radio" name="mark" value="pr">"."Project<br>";
echo "<input type="Radio" name="mark" value="fi">"."Final<br>";
I looked for this problem in your website and other websites but I couldn't find the solution for it. 我在您的网站和其他网站中寻找了此问题，但找不到解决方案。
3 个解决方案
解决方案1
7 2011-10-29 17:49:50
Here your syntax is off. 在这里，您的语法已关闭。 use single quotes like you did in your form tag, or escape the double quotes 像在表单标签中一样使用单引号，或者转义双引号
echo "<input type=\"Radio\" name=\"mark\" value=\"mt\">"."MidTerm<br>";
...
...
and you need to fix this line too, here: 您也需要在以下位置修复此行：
echo "<table border=\"3\" cellpadding=\"3\">";
also, you can get by without, but you should quote your size values too towards the bottom 同样，您也可以没有，但是您也应该在底部引用size值
解决方案2
2 2011-10-29 17:49:34
echo "<input type="Radio" name="mark" value="mt">"."MidTerm<br>";
Escape the Double quotes 转义双引号
echo "<input type=\\"Radio\\" name=\\"mark\\" value=\\"mt\\">"."MidTerm<br>";
similarly for other lines. 其他行也一样。
====== ======
alternatively use single quotes in one of the places. 或者在其中一个地方使用单引号。
echo '<input type="Radio" name="mark" value="mt">'.'MidTerm<br>';
OR 要么
echo "<input type='Radio' name='mark' value='mt'>"."MidTerm<br>";
解决方案3
0 2011-10-29 17:47:56
Why do you think you have an error? 为什么您认为自己有错误？ You're not escaping quotes in the string, of course it's not going to work. 您不是在转义字符串中的引号，当然这是行不通的。 Just add a \\ before the " in the <input> tag and you're all good. 只需在<input>标记中的"之前添加一个\\ ，就可以了。
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