sql retrieve data from table and make them links
Question
I am generating web pages from database. Now my question is:
I have 1000 records(names) in my database(MySql). I have made a search box in a page and when i enter any name or a part of name that is in my DB all the name's should come up. Eg-
SELECT * FROM table where name like '%$find%'
Now i want to show the selected names(fetched through the query) on the new page so that when i click on any of the name a new page should open up and all the data related to that selected name (present in the table belonging to the database)to be shown on that page with navigation buttons, what query should i use to perform it.
In short i want to make my page like Google search page.
My first page is like this
<html>
<body >
<h2>Search</h2>
<form name="search" method="post" action="second.php">
Search Name: <input type="text" name="find" id="find" />
<input type="submit" name="search" value="search" />
</form>
</body>
</html>
Second page is somewhat like this
<html>
<head>
<script>
function favBrowser()
{
var mylist=document.getElementById("opt");
document.getElementById("favorite").value=mylist.options[mylist.selectedIndex].text;
}
</script>
</head>
<body>
<form method="get">
<?php
$find = $_REQUEST['find'];
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("data", $con);
$result = mysql_query("SELECT * FROM table where name like '%$find%'");
$result_rows = mysql_num_rows($result);
while($row = mysql_fetch_array($result))
{
// $names[] = $row['name'];
// echo $names[0];
// echo "$row[name]. $row[id] <a href='data.php?edit=$row[name]'>edit</a><br />";
$_name = $row['name'];
echo "Name : <input type='text' name='name' value='$_name' size='30'>";
echo "<br />";
}
}
mysql_close($con);
?>
<!--</select>
<input type ="submit" value="submit">
<p>Your selected name is: <input type="hidden" name="fun" id="favorite" size="30">
</p>
-->
</body>
</html>
2 answers
solution1
1 ACCPTED 2012-12-13 10:46:45
Well, simplified, on the first page you'll have something like:
while($row = mysql_fetch_array($result))
{
$_name = $row['name'];
echo '<a href="second_page.php?name='.strip_tags($_name)'" target="_BLANK"'.'</a>';
}
and on the second page you have name, passed as URL parameter, on which you then do another database look up to get the contacts details and populate the various fields:
$_name = $GET['name'];
Please remember to add the required escapes or rather use PDO / mysqli
solution2
0 2012-12-13 11:17:49
但问题是如何将所有名称作为链接，然后在下一页上获取结果..对吗？
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