Query and combobox

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lid
Forum Newbie
Posts: 1
Joined: Wed Mar 19, 2008 7:18 am

Query and combobox

Post by lid »

HI!

I have yet problem with combobox and query.


COMBOBOX PAGE


This page work, I have just one problem:
With this code : " $sql ="select name from [db]..sysobjects where
xtype = 'U'order by
name";" in my combobox appear the name of all the tables of the db
but I want only few tables. how i can change this code?
//-------------------------------------------------------------------------­­---------------------------------------

Code: Select all

 
<?php 
error_reporting(E_ALL); 
ini_set('display_errors', 1); 
 
 
$dbname = "[aa]"; 
$link = mssql_connect("aa", "ll", "ll")or die("Could not connect: " . 
mssql_error()); 
mssql_select_db($dbname) or die('Could not select database'); 
?> 
 
 
<html> 
<head></head> 
<body> 
<?php 
 
 
// SELECT FUNZIONANTE MA ESTRAE TUTTI I CAMPI 
$sql ="select name from [db]..sysobjects where xtype = 'U'order by 
name"; 
 
 
// processi la stringa $result = mssql_query($sql); 
 
 
//verifichi il risultato se diverso da 0. Se fallisce significa che 
non ci sono tabelle nel db 
if (!$result) { 
echo "DB Error, could not list tables\n"; 
echo ('MsSQL Error: ' . mssql_error()); 
 
 
exit; 
 
 
 
} 
 
 
?> 
 
<table width="100%" border="0" cellpadding="0" cellspacing="0"> 
<tr> 
<td width="171">Select table </td> 
<td width="620">codice</td> 
</tr> 
<tr> 
<td><form name="myform" method="post" action="page2.php"> 
Tables: <select name="category" onChange="submit()"> 
<option value=""></option> 
 
 
<?php 
 
 
// cicli per estrarre tutti i nomi delle tabelle e crei direttamente 
il form di scelta. 
while ($row = mssql_fetch_row($result)) { 
echo "<option value=\"".$row[0]."\">".$row[0]."</option>"; 
 
 
 
} 
 
 
mssql_free_result($result); 
?> 
 
</select> 
</form> 
</td><td>&nbsp;</td> 
</tr> 
<tr><td>&nbsp;</td><td>&nbsp;</td> 
</tr> 
</table> 
 
 
</body> 
</html> 
 
-------------------------------------------------------------------------


PAGE 2.PHP


-----------------------------

Code: Select all

<?php 
 
 
error_reporting(E_ALL); 
ini_set('display_errors', 1); 
 
 
$dbname = '[DB]'; 
$link = mssql_connect(dd, aa, aa)or die("Could not connect: " . 
mssql_error()); 
mssql_select_db($dbname) or die('Could not select database'); 
 
 
if(isset($_POST['nome'])) { 
$azienda = $_POST['nome']; 
if($azienda == '[Telephony Daily A]'){ 
$sql = "SELECT campo1, campo2, campo3, campo4, campo5  from 
$azienda"; 
 
 
 
}else if($azienda == "[Telephony Daily B]"){ 
 
 
$sql = "SELECT campo1, campo3, campo7, campo9, campo11  from 
$azienda"; 
 
 
}else{ 
 
 
echo "errore nella definizione del nome dell'azienda'"; 
 
 
 
} 
echo $sql; 
} 
 
 
?> 
 
<html> 
<head> 
</head> 
<body> 
 
 
<?php 
 
 
$result = mssql_query($sql); 
 
 
//verifichi il risultato se diverso da 0. Se fallisce significa che 
non ci sono tabelle nel db 
if (!$result) { 
echo "DB Error, could not list tables\n"; 
//echo 'MsSQL Error: ' . mssql_error(); 
exit; 
 
 
 
} 
 
 
?> 
 
<table width="100%" border="1" cellpadding="0" cellspacing="0"> 
<tr><td><div align="center">name/address</div></td><td><div 
align="center">department/surname</div></td><td><div 
align="center">field/history</div></td></tr> 
<?php 
 
 
// cicli l'array dei record per ircavare i dati 
while($row = mssql_fetch_row($result)){ 
echo "<tr><td><div>s".$row[0]."</div></td><td><div>".$row[1]."</ 
div></ 
td><td><div> 
".$row[2]."</div></td></tr>"; 
 
 
 
} 
 
 
mssql_free_result($result); 
msysql_close($link); 
?> 
 
</table> 
</body> 
</html> 
 
 
?> 
 
---------------------------------------------------------------------------­­----


Those Are The Errors for the page2.php


Notice: Undefined variable: sql in C:\Program Files\xampp\htdocs
\intranet\page2.php on line 35


Warning: mssql_query() [function.mssql-query]: Query failed in C:
\Program Files\xampp\htdocs\intranet\page2.php on line 35
DB Error, could not list tables


The line 35 is this: $result = mssql_query($sql);


Thanks Lidia
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