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tokkille

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Posts posted by tokkille

  1. Thanks less then three!!

    My new code looks like this:

    <html>
    <head>
    <title>G&ouml;tt 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">S&ouml;k vem som &auml;r g&ouml;ttigast </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
  2. 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>G&ouml;tt mos</title>
    <link href="../skolan.css" rel="stylesheet" type="text/css">
    </head>

    <body>
    <h3 align="center">Goa Grejer </h3>
    <p align="center">S&ouml;k vem som &auml;r g&ouml;ttigast </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
  3. 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
  4. 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
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