当用户输入ID时，在mysql表中显示行
[英]Display row in mysql table when a user enter id
问题描述
I want to display a row when a user enter his id and after submit. 
我想在用户输入ID并提交后显示一行。 My code is not working correctly. 我的代码无法正常工作。 Please rectify this. 请纠正这一点。
search-form.php is looking like this. search-form.php看起来像这样。
</html><body>
<form method="GET" action="search.php">
Keyfield <input type="text" name="search"> <br><br>
<input type="submit" value="submit">
</form></body>
</html>
and 和
search.php looking like this. search.php看起来像这样。
<?php
$connection = mysql_connect('localhost','user','pass') or die ("Couldn't connect to server.");
$db = mysql_select_db('db', $connection) or die ("Couldn't select database.");
$search=$_GET['search'];
$fetch = 'SELECT * FROM `table` WHERE `ID` = "'.$search.'"';
echo "<table margin=auto width=999px border=1>";
echo "<tr><td><b>ID</b></td><td><b>Name</b></td><td><b>Telephone</b></td><td> <b>E-mail</b></td><td><b>Couttry Applying for</b></td><td><b>Visa-Category</b> </td><td><b>Other Category</b></td><td><b>Passport No</b></td><td> <b>Remarks</b></td></tr>";
for($i=0;$i<$num;$i++)
{
$row=mysql_fetch_row($fetch);
echo "<tr>";
echo"<td>$row[0]</td>";
echo"<td>$row[1]</td>";
echo"<td>$row[2]</td>";
echo"<td>$row[3]</td>";
echo"<td>$row[4]</td>";
echo"<td>$row[5]</td>";
echo"<td>$row[6]</td>";
echo"<td>$row[7]</td>";
echo"<td>$row[8]</td>";
echo"</tr>";
}//for
echo"</table>";
?>
Display when a user enter his id and submit. 当用户输入其ID并提交时显示。 But this code doesn't display row with id. 但是此代码不显示带有id的行。 Rectify this. 纠正这个。 Thanks. 谢谢。
2 个解决方案
解决方案1
0 2013-06-26 09:36:00
you are missing mysql_query statement.You should execute the query before fetching the result modify like $sql= 'SELECT * FROM table WHERE ID = "'.$search.'"'; 您缺少mysql_query语句。应在获取结果修改之前执行查询，例如$ sql ='SELECT * FROM table WHERE ID =“'。$ search。'”'; $fetch = mysql_query($sql); $ fetch = mysql_query（$ sql）;
解决方案2
0 2013-06-26 09:38:41
A few points: 几点：
- mysql_connect is obsolete, do not use it. mysql_connect已过时，请勿使用。 Use PDO instead for example. 例如，使用PDO。
-
$fetch = 'SELECT * FROMtableWHEREID= "'.$search.'"'leads to the most common and severe security flaw : SQL injection.$fetch = 'SELECT * FROMtableWHEREID= "'.$search.'"'导致最常见和最严重的安全漏洞：SQL注入。 Please read about this (google)请阅读有关此内容（谷歌） - Where do you "fetch" the result of your query ? 您在哪里“获取”查询结果？
About point 3, assuming the fact that you will use PDO, please read http://www.phpro.org/tutorials/Introduction-to-PHP-PDO.html 关于第3点，假设您将使用PDO，请阅读http://www.phpro.org/tutorials/Introduction-to-PHP-PDO.html
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