a href and php combination

[cilian29]

hi all I'm trying to make this work, can you help on my script i'm doing now I'm trying to call my php link within functions. pls bear with me bec. I'm lost my my self. im trying to call a list of items from my database when I clicked the employee nos.by the way I'm using posrgesql and apache server and its running good.

 

 

I'm getting this error

 

 

Warning: pg_query() [function.pg-query]: Query failed: ERROR: syntax error at or near ")" at character 44 in C:\Program Files\sample.php

 

Warning: pg_fetch_array() expects parameter 1 to be resource, string given in C:\Program Files\sample.php

 

I have a query function  called

 

function callme(){

$sql2 =" SELECT * FROM employee WHERE empid = ($myid)";

$result=pg_query($sql2);

//or die(pg_last_error());

 

while($row=pg_fetch_array($sql2)){

 

echo $row[empid];

}

Then I create a swich case to call my function

 

switch($_REQUEST["num"])

{

case "1":

 

callme();

break;

case "2":

 

echo 'no record!!';

break;

 

default:

break;

}

 

 

}

 

 

this is my php file:

 

$mod = $_GET[mod];

$myid =$_GET[id];

$sq1=pg_query("SELECT DISTINCT empid FROM employee

INNER JOIN client c ON(merchantable.ccindex = c.clientindex)

ORDER BY merindex ASC");  :shy:this code is running its showing me what distinct items i need.

 

while ($row = pg_fetch_array($sq1)){

echo '<table>';

echo '<tr>';

echo "<td><a href='?mod=$mod&num=1&id=$myid& $row[refid]>".$row["refid"]."</a></td>";

echo "</tr>";

echo '</table>';

}

then my url :http://localhost/preprod/overwriteA.php?mod=&num=1&id=&%201004SA%3E1004SA%3C/a%3E%3C/td%3E%3Ctable%3E%3Ctr%3E%3Ctd%3E%3Ca%20href=

 

Why am I not getting anywhere PLEase HELp

[btherl]

Inside your function callme(), $myid is not defined.  You can pass it as an argument like this:

 

callme($myid);

function callme($myid) {
   ...
}

 

A good debugging technique for this situation is to print out the query which fails.  Then you would see that $myid is missing.

[cilian29]

Thank you for your quick reply and I hope you help me further when you mention pass the callme($myid) as an argument heres what I did. I just added the call me above my original function.

I created anew php file to test on it sample.php

 

$callme($myid);

 

function callme($myid){

 

$sql2 =" SELECT * FROM employee WHERE empid = ($myid)";

      $result=pg_query($sql2);

            //or die(pg_last_error());

     

        while($row=pg_fetch_array($sql2)){

       

        echo $row[empid];

        }

 

}

 

my question is, do I remove my switch case when I call the function? when I run the my code above this time I dont see any errors but there is no display on my browser and my

url: http://localhost/preprod/sample.php?mod=&nir=1&id=&%201004SA>1004SA</a></td><table><tr><td><a%20href=http://localhost/preprod/sample.php?mod=&num=1&id=&%201004SA>1004SA</a></td><table><tr><td><a%20href=

 

[Jessica]

Please use code tags.

 

When you call it in the switch statement, you'll need to pass the argument.

[btherl]

If that is your whole sample.php, a few things are missing.  Try this:

 

$myid =$_GET[id];

/* Here you need the call to pg_connect(), so you can call pg_query() on that connection */

callme($myid); /* Don't put "$" in front of callme */

function callme($myid){

   $sql2 =" SELECT * FROM employee WHERE empid = ($myid)"; 
         $result=pg_query($sql2);
            or die(pg_last_error());  /* Don't comment out the "or die" - you need it to find the bug */
      
         while($row=pg_fetch_array($result)){  /* Use pg_fetch_array() on $result, not on $sql2 */
         
         echo $row[empid];
         }

}

[cilian29]

good day thank you for all helping me and for the replies

on seeing the code you suggested I immediately tried adding the changes but unfortunately I came upon one problem to the next so here it goes and I made some difference on my previous script above.

 

Parse error: parse error, unexpected T_LOGICAL_OR in C:\Program Files\xampp\htdocs\preprod\overwriteB.php on line 40

 

im getting this error when the code is inserted so i comment it out for a while.

or die(pg_last_error());  /* Don't comment out the "or die" - you need it to find the bug */

 

Warning: pg_query() [function.pg-query]: Query failed: ERROR: syntax error at or near ")" at character 35 in C:\Program Files\xampp\htdocs\preprod\sample.php on line 39

 

Warning: pg_fetch_array() expects parameter 1 to be resource, boolean given in C:\Program Files\xampp\htdocs\preprod\sample.php on line 41

 

1 <?php

2 include('server/configA.php');

3

4 $mod = $_GET[mod];

5 $myid =$_GET[id];

6

7 $link=pg_connect("host= " .DB_HOST. " port= " .DB_PORT. " dbname= " .DB_DATABASE. " user= " .DB_USER. " password= " .DB_PSWD);

8 if(!$link)

9 {

10 die('Failed to connect to server: ' .pg_result_error_field($link));

11 }

12 callme($myid); /* Don't put "$" in front of callme */

13 if($_REQUEST[nir]==1){callme();}

14

15

16

17

18 $sq1 ="SELECT * FROM merchantable";

19 $result =pg_query($sq1);

20

21 //for refid display below

22 while ($row = pg_fetch_array($result)){

23 echo '<table>';

24 echo '<tr>';

25

26

27 echo "<td><a href='?mod=$mod&nir=1&id=$myid& $row[empid]'>".$row["empid"]."</a></td>";

28

29 echo '</tr>';

30 echo '</table>';

31 }

32

33

34

35  function callme($myid){

36

37 $q1 =" SELECT * FROM nir WHERE empid = ($myid)";

38 //$q1 ="SELECT * FROM nir";

39 $result =pg_query($q1);

40 //or die(pg_last_error());  /* Don't comment out the "or die" - you need it to find the bug */

41     

42

43 while($rows=pg_fetch_array($result)){  /* Use pg_fetch_array() on $result, not on $q1 */

44       

45        echo $rows[empid];

46        }

47 }

48 ?>

 

thank you again hope you van help me... pls!

 

 

 

[btherl]

Change this:

 

 $result =pg_query($q1);
//or die(pg_last_error());  /* Don't comment out the "or die" - you need it to find the bug */

 

to this:

 

 $result =pg_query($q1) /* Remove ";" here when uncommenting the next line */
or die(pg_last_error());  /* Don't comment out the "or die" - you need it to find the bug */

 

Then the next thing to do is to print out the query which has the error.  Add this line:

 

echo "<br>Query is $q1<br>";

 

Add that after $q1 has been set, but BEFORE it use used in pg_query().  Then post what it prints here.