Forum

Scris: **Dum Oct 03, 2010**

Buna am si eu o problema cu urmatorul cod:

<?php
$cuvant=$_GET['q'];
if($cuvant == NULL || $cuvant == '') {echo "<br>Sorry but you did not write any word";}

$conn = mysql_connect("localhost", "cautari", "cautari") 
    or die("Unable to connect to MySQL"); 
$selectdb = mysql_select_db('cautari'); 
$interogare = "SELECT * FROM siteuri_en WHERE cuvinte_cheie LIKE '%$cuvant%' OR descriere_site LIKE '%$cuvant%' "; 
$rezultat = mysql_query($interogare); 
if (mysql_errno()) { 
    echo "<br />". mysql_errno(). " : ". mysql_error(). "<br />"; 
} 
$nr_campuri = mysql_num_fields($rezultat); 
if (mysql_num_rows($rezultat) == 0) { 
    echo '<br><br>Your search - <b>'.$cuvant.'</b> - did not match any result.<br><br>

Suggestions:<br><br>

*Make sure all words are spelled correctly.<br>
*Try different keywords.<br>
*Try more general keywords<br>';
} 
else { 
    while ($rand = mysql_fetch_assoc($rezultat)) {    


echo '<br><table>
    <tr>
    <td></td><td align="left"><font size="5"><a href="'.$rand['link_site'].'" target="_blank">'.$rand['nume_site'].'</a></font><br><font color="<?php $ip = $_SERVER['REMOTE_ADDR'];
$conn = mysql_connect("localhost", "cautari", "cautari")
    or die("Unable to connect to MySQL");
$selectdb = mysql_select_db('cautari');
$utilizator=mysql_real_escape_string($_GET['utilizator']);
$cerereSQL = 'SELECT * FROM design WHERE ip="'.$ip.'"'; 
$rezultat = mysql_query($cerereSQL);
if(mysql_num_rows($rezultat) > 0) {
while($rand = mysql_fetch_array($rezultat))	{ 
      echo ''.$rand['culoare_descriere'].'';
  }
}?>" size="4">'. str_replace($cuvant, '<u><b>'.$cuvant.'</b></u>', $rand['descriere_site']). '</font><br><font color="red" size="4">'.$rand['link_site'].'</font><br></td>
    </tr><tr>
    <td><font size="1pt"></font></td><td align="right"></td>
    </tr><tr>
    <td align="center" colspan="2">

    </tr>
    </table>';
  }
}
?>

Problema incepe de la:

Cod: Selectaţi tot

<font color="<?php $ip = $_SERVER['REMOTE_ADDR'];
$conn = mysql_connect("localhost", "cautari", "cautari")
    or die("Unable to connect to MySQL");
$selectdb = mysql_select_db('cautari');
$utilizator=mysql_real_escape_string($_GET['utilizator']);
$cerereSQL = 'SELECT * FROM design WHERE ip="'.$ip.'"'; 
$rezultat = mysql_query($cerereSQL);
if(mysql_num_rows($rezultat) > 0) {
while($rand = mysql_fetch_array($rezultat))	{ 
      echo ''.$rand['culoare_descriere'].'';
  }
}?>" size="4">

Eu as vrea ca sa arate color din baza de date care o seteaza vizitatorul dar el imi da erroare cand intru pe site: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/hosting/gamestime/site-search.co.cc/search.php on line 86

Pls cum as putea sa fk sa arate culoarea din baza de date??ms

Scris: **Dum Oct 03, 2010**

Codul de sus e varza, cu o gramada de greseli. nu ai ce corecta la el ci trebuie regandit /refacut.
De ce de 2 ori conectare la mysql in acelasi fisier?
A doua conectare intr-o bucla WHILE.
Deschidere <?php intr-un echo (deci era deja in structura <?php ... ?>)
Conectare, Select si parcurgere intr-un echo deschis.

Deci, cine l-a facut sa-l refaca.

Scris: **Dum Oct 03, 2010**

Acel cod eu lam facut. Nu prea stiu asa de multe despre php dar am o intrebare pot sa fk doua cereri de mysql in acelasi cod php?? de ex asa:

Cod: Selectaţi tot

<?php
$cuvant=$_GET['q'];


if($cuvant == NULL || $cuvant == '') {echo "<br>Sorry but you did not write any word";}

$conn = mysql_connect("localhost", "cautari", "cautari") 
    or die("Unable to connect to MySQL"); 
$selectdb = mysql_select_db('cautari'); 
$interogare = "SELECT * FROM siteuri_en WHERE cuvinte_cheie LIKE '%$cuvant%' OR descriere_site LIKE '%$cuvant%' "; 
$rezultat = mysql_query($interogare); 
$cerereSQL = 'SELECT * FROM design WHERE ip="'.$ip.'"'; 
$rezultat2 = mysql_query($cerereSQL);
if(mysql_num_rows($rezultat2) > 0) {
while($rand = mysql_fetch_array($rezultat2))	{ 
if (mysql_errno()) { 
    echo "<br />". mysql_errno(). " : ". mysql_error(). "<br />"; 
} 
$nr_campuri = mysql_num_fields($rezultat); 
if (mysql_num_rows($rezultat) == 0) { 
    echo '<br><br>Your search - <b>'.$cuvant.'</b> - did not match any result.<br><br>

Suggestions:<br><br>

*Make sure all words are spelled correctly.<br>
*Try different keywords.<br>
*Try more general keywords<br>';
} 
else { 
    while ($rand = mysql_fetch_assoc($rezultat)) {    


echo '<br><table>
    <tr>
    <td></td><td align="left"><font size="5"><a href="'.$rand['link_site'].'" target="_blank">'.$rand['nume_site'].'</a></font><br><font color="'.$rand['culoare_descriere'].'" size="4">'. str_replace($cuvant, '<u><b>'.$cuvant.'</b></u>', $rand['descriere_site']). '</font><br><font color="red" size="4">'.$rand['link_site'].'</font><br></td>
    </tr><tr>
    <td><font size="1pt"></font></td><td align="right"></td>
    </tr><tr>
    <td align="center" colspan="2">

    </tr>
    </table>';
  }
}
}
?>

ThX

Scris: **Lun Oct 04, 2010**

Da, pot fi facute mai multe cereri, interogari SQL in acelasi cod.
Ceea ce nu stiti daca se poate, legat de php, incercati si vedeti rezultatul.

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Problema afisare date din db

Problema afisare date din db

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