Sorrow Posted November 22, 2006 Share Posted November 22, 2006 I am Displaying all those variable properly until i get to ( echo "<td>" . $row['Adresse'] . "</td></tr>";) it is diplaying (echo "<tr><td>Adresse : </td>) but not the database value.if you wanna see here is the website : [url=http://www.pirate-production.com/webrencontre/profil.php]http://www.pirate-production.com/webrencontre/profil.php[/url][code]<?php$con = mysql_connect("","","");if (!$con) { die('Could not connect: ' . mysql_error()); }mysql_select_db("jplegris_pirate", $con);$result = mysql_query("SELECT * FROM Usagers");while($row = mysql_fetch_array($result)) {echo "<table border='0'><tr><td>Nom Usager : </td>"; echo "<td>" . $row['Username'] . "</td></tr>"; echo "<tr><td>Nom : </td>"; echo "<td>" . $row['Nom'] . "</td></tr>"; echo "<tr><td>Prenom : </td>"; echo "<td>" . $row['Prenom'] . "</td></tr>"; echo "<tr><td>Âge : </td>"; echo "<td>" . $row['Age'] . "</td></tr>"; } echo "<tr><td>Adresse : </td>"; echo "<td>" . $row['Adresse'] . "</td></tr>";echo "</table>";?>[/code] Quote Link to comment Share on other sites More sharing options...
DaviDJ Posted November 22, 2006 Share Posted November 22, 2006 edit out your mysql_connect details, and also is that the correctly spelt field name?Double check it is and also double check there is a value set for the row(s) you are looking at Quote Link to comment Share on other sites More sharing options...
hitman6003 Posted November 22, 2006 Share Posted November 22, 2006 you also have a lot of extra echo statements that are unnecessary.[code]$con = mysql_connect(***, ***, ***) or die(mysql_error());mysql_select_db("database", $con);$result = mysql_query("SELECT * FROM Usagers");while($row = mysql_fetch_array($result)) { echo " <table border='0'> <tr> <td>Nom Usager : </td> <td>" . $row['Username'] . "</td> </tr> <tr> <td>Nom : </td> <td>" . $row['Nom'] . "</td> </tr> <tr> <td>Prenom : </td> <td>" . $row['Prenom'] . "</td> </tr> <tr> <td>Age : </td> <td>" . $row['Age'] . "</td> </tr> <tr> <td>Adresse : </td> <td>" . $row['Adresse'] . "</td> </tr> </table>";}[/code] Quote Link to comment Share on other sites More sharing options...
Sorrow Posted November 22, 2006 Author Share Posted November 22, 2006 ty very much it is working now but is there a way to use a $_POST['Username'] in a select query like[code]$result = mysql_query("SELECT * FROM Usagers where Username = "$_Post['Username']);[/code] Quote Link to comment Share on other sites More sharing options...
hitman6003 Posted November 22, 2006 Share Posted November 22, 2006 Use the concatenation operator ( . - a period):[code]$result = mysql_query("SELECT * FROM Usagers where Username = " . $_Post['Username']);[/code] Quote Link to comment Share on other sites More sharing options...
Sorrow Posted November 22, 2006 Author Share Posted November 22, 2006 It still gives me an error : mysql_fetch_array(): supplied argument is not a valid MySQL result resource in [code]mysql_select_db("jplegris_pirate", $con);$result = mysql_query("SELECT * FROM Usagers where Username = " . $_Post['Username']);while($row = mysql_fetch_array($result)) { echo " <table border='0' align='center'> <tr> <td>Nom Usager : </td> <td>" . $row['Username'] . "</td> </tr> <tr> <td>Nom : </td> <td>" . $row['Nom'] . "</td> </tr> <tr> <td>Prenom : </td> <td>" . $row['Prenom'] . "</td> </tr> <tr> <td>Age : </td> <td>" . $row['Age'] . "</td> </tr> <tr> <td>Sexe : </td> <td>" . $row['Sexe'] . "</td> </tr> <tr> <td>Adresse : </td> <td>" . $row['Adresse'] . "</td> </tr> <tr> <td>Ville: </td> <td>" . $row['Ville'] . "</td> </tr> <tr> <td>Province: </td> <td>" . $row['Province'] . "</td> </tr> <tr> <td>Pays: </td> <td>" . $row['Pays'] . "</td> </tr> <tr> <td>Adresse : </td> <td>" . $row['Adresse'] . "</td> </tr> <tr> <td>No. de Tel. : </td> <td>" . $row['NoTel'] . "</td> </tr> <tr> <td>Taille Physique : </td> <td>" . $row['TaillePhysique'] . "</td> </tr> <tr> <td>Fumeur : </td> <td>" . $row['Fumeur'] . "</td> </tr> <tr> <td>Buveur : </td> <td>" . $row['Buveur'] . "</td> </tr> <tr> <td>Ethnie : </td> <td>" . $row['Ethnie'] . "</td> </tr> <tr> <td>Courriel : </td> <td>" . $row['Courriel'] . "</td> </tr> </table>";}?>[/code] Quote Link to comment Share on other sites More sharing options...
hitman6003 Posted November 22, 2006 Share Posted November 22, 2006 It's a string so you have to put it in single quotes:[code]$result = mysql_query("SELECT * FROM Usagers where Username = '" . $_Post['Username']) . "'";[/code] Quote Link to comment Share on other sites More sharing options...
Sorrow Posted November 22, 2006 Author Share Posted November 22, 2006 Ive edited the string a little [code]$result = mysql_query("SELECT * FROM Usagers where Username = '" . $_Post['Username']. "'") ;[/code]but still not working Quote Link to comment Share on other sites More sharing options...
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