PHP SQL搜索页面返回一个错误 [英] PHP SQL Search Page returns an error
问题描述
< html>
< head>
< title>搜寻< / title>
< style type =text / css>
table {
background-color：#FCF;
}
th {
width：150px;
text-align：left;
}
< / style>
< / head>
< body>
< h1>搜寻< / h1>
< form method =postaction =search.php>
< input type =hiddenname =submittedvalue =true/>
< label>搜索|类别：
< select name =category>
< option value =fname> FName< / option>
< option value =lname> LName< / option>
< / select>
< / label>
< label>搜索条件：< input type =textname =criteria/>< / label>
< input type =submit/>
< / form>
$ b if（isset（$ _ POST ['submitted']））{
//连接到DB
包括（'connect.php'）;
$ category = $ _POST ['category'];
$ criteria = $ _POST ['criteria'];
$ query =SELECT * FROM people WHERE category ='$ category';
$ result = mysqli_query（$ dbcon，$ query）或die（'Error'）;
echo< table>;
回显< thead>;
回显< tr>;
echo< th> First Name< / th>;
echo< th> Last Name< / th>;
回声< / tr>;
回显< / thead>;
回声< tbody>;
if（mysqli_num_rows（$ result））{
while（$ row = mysqli_fetch_array（$ result，MYSQLI_ASSOC））{
echo< tr>;
echoth 。 $ row ['fname']。 < /第> 中;
echoth 。 $ row ['lname']。 < /第> 中;
回声< / tr>;
}
}
echo< / tbody>;
echo< / table>;
}
？>
< / body>
< / html>
可能因为这一行： $ result = mysqli_query（$ dbcon，$ query）或die（'Error'）;
如果查询失败，脚本将停止并打印Error。<
您可能希望将 mysqli_error（$ dbcon）而不是'Error'和在这种特定情况下，我建议回显查询以了解它的外观。
请在数据库查询中使用POST数据之前转义POST数据，或者使用预先准备的语句！
Okay guys, maybe I can get some definitive answers for this newest issue. This PHP section doesn't return any responses, however, I know the data is there. Does anyone know why this returns the text "Error"?
<html>
<head>
<title>Search</title>
<style type="text/css">
table {
background-color: #FCF;
}
th {
width: 150px;
text-align: left;
}
</style>
</head>
<body>
<h1>Search</h1>
<form method="post" action="search.php">
<input type="hidden" name="submitted" value="true"/>
<label> Search | Category:
<select name="category">
<option value="fname">FName</option>
<option value="lname">LName</option>
</select>
</label>
<label>Search Criteria: <input type="text" name="criteria"/></label>
<input type="submit"/>
</form>
<?php
if (isset($_POST['submitted'])) {
// connect to DB
include('connect.php');
$category = $_POST['category'];
$criteria = $_POST['criteria'];
$query = "SELECT * FROM people WHERE category = '$category'";
$result = mysqli_query($dbcon, $query) or die ('Error');
echo "<table>";
echo "<thead>";
echo "<tr>";
echo "<th>First Name</th>";
echo "<th>Last Name</th>";
echo "</tr>";
echo "</thead>";
echo "<tbody>";
if (mysqli_num_rows($result)) {
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
echo "<tr>";
echo "<th>" . $row['fname'] . "</th>";
echo "<th>" . $row['lname'] . "</th>";
echo "</tr>";
}
}
echo "</tbody>";
echo "</table>";
}
?>
</body>
</html>
Probably because of this line: $result = mysqli_query($dbcon, $query) or die ('Error');
If the query fails, the script will stop and print "Error".
You might want to put something like mysqli_error($dbcon) instead of 'Error' and in this specific case, I would recommend echoing out the query to see how it looks.
And please, either escape the POST data before using it in a database query, or rather use prepared statements!
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