tokkille
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Posts posted by tokkille
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Thanks less then three!!
My new code looks like this:
<html>
<head>
<title>Gött mos</title>
<link href="file:///C|/Documents and Settings/vahlne/Desktop/PHP/Kap48/skolan.css" rel="stylesheet" type="text/css">
</head>
<body>
<h3 align="center">Goa Grejer </h3>
<p align="center">Sök vem som är göttigast </p>
<hr>
<?php
if (isset($_POST["searchstring"]))
{
$searchstring = $_POST["searchstring"]
$searchtype = $_POST["searchtype"]
$sql="SELECT * FROM maraton1 WHERE $searchtype LIKE '%$searchstring%' ORDER BY Plats ASC";
include("XXX.php");
$result = mysql_query($sql,$db);
echo "<table border=1 align=center cellspacing=2 cellpadding=3>\n";
echo "<TR><TH>Plats<TH>Person<TH>Poäng</TR>\n";
while ($rad = mysql_fetch_array($result))
{
echo "<TR><TD>$rad[Plats]<TD>$rad[Person]<TD>$rad[Poäng]\n";
}
echo "</TABLE>";
}
else
{
?>
<form method="POST" action="<?php $PHP_SELF ?>">
<table border="1" align="center" cellspacing=2 cellpadding=3>
<tr><td>Sök här</td>
<td>Söktyp</td></tr>
<tr>
<td><input type="text" name="searchstring" size="40"></td>
<td><select size="1" name="searchtype">
<option selected value="Person">Person</option>
<option value="Plats">Plats</option>
</select></td>
</tr>
</table>
<p align="center">
<input type="submit" value="Sök" name="B1">
<input type="reset" value="Töm sökraden ovan" name="B2">
</p>
</form>
<?php
}
?>
<hr>
<p align="center"><a href="index.php">Tillbaks</a>
<p>
</body>
</html>
But know it complains about:
Parse error: parse error, unexpected T_VARIABLE in /customers/vahlne.se/vahlne.se/httpd.www/sok funktion.php on line 17
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Hi,
I have maneged to recive information from my database. but when i put in a search function on the homepage to search the database it dosent retrive the result in a table?!
This is what it looks like:
<html>
<head>
<title>Gött mos</title>
<link href="../skolan.css" rel="stylesheet" type="text/css">
</head>
<body>
<h3 align="center">Goa Grejer </h3>
<p align="center">Sök vem som är göttigast </p>
<hr>
<?php
if ($searchstring)
{
$sql="SELECT * FROM maraton1 WHERE $searchtype LIKE '%$searchstring%' ORDER BY Plats ASC";
include("XXX.php");
$result = mysql_query($sql,$db);
echo "<table border=1 align=center cellspacing=2 cellpadding=3>\n";
echo "<TR><TH>Plats<TH>Person<TH>Poäng</TR>\n";
while ($rad = mysql_fetch_array($result))
{
echo "<TR><TD>$rad[Plats]<TD>$rad[Person]<TD>$rad[Poäng]\n";
}
echo "</TABLE>";
}
else
{
?>
<form method="POST" action="<?php $PHP_SELF ?>">
<table border="1" align="center" cellspacing=2 cellpadding=3>
<tr><td>Sök här</td>
<td>Söktyp</td></tr>
<tr>
<td><input type="text" name="searchstring" size="40"></td>
<td><select size="1" name="searchtype">
<option selected value="Person">Person</option>
<option value="Plats">Plats</option>
</select></td>
</tr>
</table>
<p align="center">
<input type="submit" value="Sök" name="B1">
<input type="reset" value="Töm sökraden ovan" name="B2">
</p>
</form>
<?php
}
?>
<hr>
<p align="center"><a href="../../../../Application Data/Macromedia/Dreamweaver 8/Configuration/ServerConnections/www.vahlne.se//index.php">Tillbaks</a>
<p>
</body>
</html>
Thanks for any help!
/Tokkille -
Thanks for the help!!!!!
/Tokkille -
anatak,
Thanks alot i fixed it!!!!!
Im super grateful!!!!
Tokkille -
When i try to recive data from my database i get the following error message (on the webpage):
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /customers/vahlne.se/vahlne.se/httpd.www/index.php on line 18
My code in PHP looks as follows (from row 13(<?php)-24(?>)):
<?php
include("XXX.php");
$result = mysql_query("SELECT * FROM maraton1 ORDER BY Plats ASC",$db);
echo "<table border=1 align=center cellspacing=2 cellpadding=3>\n";
echo "<TR><TH>Plats<TH>Lag<TH>Poäng</TR>\n";
while ($rad = mysql_fetch_array($result))
{
echo "<TR><TD>$rad[Plats]<TD>$rad[Lag]<TD>$rad[Poäng]\n";
}
echo "</TABLE>";
?>
Would be greatful for all help!!
Thanks alot,
Tokkille
-
When i try to recive data from my database i get the following error message (on the webpage):
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /customers/vahlne.se/vahlne.se/httpd.www/index.php on line 18
My code in PHP looks as follows (from row 13(<?php)-24(?>)):
<?php
include("XXX.php");
$result = mysql_query("SELECT * FROM maraton1 ORDER BY Plats ASC",$db);
echo "<table border=1 align=center cellspacing=2 cellpadding=3>\n";
echo "<TR><TH>Plats<TH>Lag<TH>Poäng</TR>\n";
while ($rad = mysql_fetch_array($result))
{
echo "<TR><TD>$rad[Plats]<TD>$rad[Lag]<TD>$rad[Poäng]\n";
}
echo "</TABLE>";
?>
Would be greatful for all help!!
Thanks alot,
Tokkille
Cant recive data from database when adding search function
in PHP Coding Help
Posted
Thanks alot it works fin now!
/Tokkille