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6年前关闭。
我正在阅读有关PHP和HTML的教程，并且该教程中使用的代码对我来说并不起作用，我想了解原因。我在http://jsfiddle.net/csmart/vkPJf/处设置了jsfiddle，以便您了解发生了什么。该代码应该与数据库连接并添加和打印表数据。似乎代码已到达打印称为“课程”的表的部分，并且失败了。连接到数据库似乎不是失败，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，应如何纠正代码？谢谢。
注意“和”，（
6年前关闭。
我正在阅读有关PHP和HTML的教程，并且该教程中使用的代码对我来说并不起作用，我想了解原因。我在http://jsfiddle.net/csmart/vkPJf/处设置了jsfiddle，以便您了解发生了什么。该代码应该与数据库连接并添加和打印表数据。似乎代码已到达打印称为“课程”的表的部分，并且失败了。连接到数据库似乎不是失败，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，应如何纠正代码？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
听到问题了
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”，（