我正在学习有关 PHP 和 HTML 的教程，但教程中使用的代码对我来说并不像演示的那样工作，我想了解原因。我已经在 http://jsfiddle.net/csmart/vkPJf/ 设置了一个 jsfiddle这样你就可以看到发生了什么。该代码应该连接数据库并添加和打印表数据。代码似乎到达了打印名为 Courses 的表的部分，但失败了。它似乎不是连接到数据库失败，因为我可以注释掉失败的代码并使用插入语句在表中创建新的数据行。 html嵌入php的方式是否有问题，代码应该如何修正？谢谢。
<!DOCTYPE html><html><head>
<script src="jquery.min.js"></script>
<script src="jquery.validate.js"></script>
<script>
$(document).ready(function() {$("form").validate();});
</script>
</head><body>
<form method="post">
cid: <input name="cid" class="required digits" maxlength="3" minlength="3"><BR>
title: <input name="title" class="required" maxlength="200"><BR>
prof: <input name="prof" maxlength="64"><BR>
cred: <input name="cred" class="required digits" maxlength="1" minlength="1"><BR>
cap: <input name="cap" class="required digits" maxlength="2" minlength="2"><BR>
<input type="submit" value="OK">
</form>
<?php
ini_set('display_errors', 'On');
$dbhost = 'XXXXXXXX';
$dbname = 'XXXXXXXX';
$dbuser = 'XXXXXXXX';
$dbpass = 'XXXXXXXX';
$mysql_handle = mysql_connect("$dbhost", "$dbname", "$dbpass", "$dbuser")
or die("Error connecting to database server");
mysql_select_db($dbname, $mysql_handle)
or die("Error selecting database: $dbname ");
$cid = array_key_exists("cid", $_REQUEST) ? $_REQUEST["cid"] : 0;
$title = array_key_exists("title", $_REQUEST) ? $_REQUEST["title"] : '';
$prof = array_key_exists("prof", $_REQUEST) ? $_REQUEST["prof"] : '';
$cred = array_key_exists("cred", $_REQUEST) ? $_REQUEST["cred"] : 0;
$cap = array_key_exists("cap", $_REQUEST) ? $_REQUEST["cap"] : 0;
if($cid <= 0) echo"";
else if (!preg_match('/^[0-9]{3}$/',$cid)) echo "Invalid cid";
else if (!preg_match('/^[0-9]{3}$/',$cred)) echo "Invalid cred";
else if (!preg_match('/^[0-9]{3}$/',$cap)) echo "Invalid cap";
else if($cid < 0) {
$rs = mysql_query("select cid from courses where cid = ".$cid);
if (mysql_numrows($rs) == 0) {
mysql_query("insert into courses(cid,cap,cred,title,prof) values("
. $cid . "," . $cap . "," . $cred
. ",'" . mysql_real_escape_string($title) . "'"
. ",'" . mysql_real_escape_string($prof) . "')"
);
}
else {
mysql_query("update courses set cap=".$cap.", cred=".$cred . ", title='". mysql_real_escape_string($title) . "'" . ", prof='". mysql_real_escape_string($prof) . "'" . " where cid=".$cid
};
}
}
$rs = mysql_query("select cid,prof,cred,cap,title from courses");
$nrows=mysql_numrows($rs);
echo "Courses<table>";
for ($i = 0; $i < $nrows; $i++) {
echo "<tr>";
echo "<td>".htmlspecialchars(mysql_result($rs,$i,"cid"))."</td>";
echo "<td>".htmlspecialchars(mysql_result($rs,$i,"title"))."</td>";
echo "<td>".htmlspecialchars(mysql_result($rs,$i,"prof"))."</td>";
echo "<td>".htmlspecialchars(mysql_result($rs,$i,"cred"))."</td>";
echo "<td>".htmlspecialchars(mysql_result($rs,$i,"cap"))."</td>";
echo "</tr>";
}
echo '</table>';
mysql_close($mysql_handle);
?>
</body></html>
最佳答案
听到问题
else if($cid < 0) {
$rs = mysql_query("select cid from courses where cid = ".$cid);
if (mysql_numrows($rs) == 0) {
mysql_query("insert into courses(cid, cap, cred, title, prof) values(" .$cid . "," .$cap . "," .$cred . ",'" . mysql_real_escape_string($title) . "'" .
","."'" . mysql_real_escape_string($prof) . "'");
}
else {
mysql_query("update courses set cap=".$cap.", cred=".$cred . ", title='". mysql_real_escape_string($title) ."'" . ", prof='". mysql_real_escape_string($prof) . "'" . " where cid=".$cid);
}
}
注意"和)，(
关于php - 在 HTML 中嵌入 PHP - 无法在表中填充和显示数据，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/14914493/