syntax error, unexpected T_VARIABLE

Question

filipgothic 0 Junior Poster in Training

10 Years Ago

I have this error upon loading page

Parse error: syntax error, unexpected T_VARIABLE in /home/a9752689/public_html/zadatak2/user.php on line 165

can you help me with it? I will post part code, if you need whole tell me

165 line is 28 here

<p align="center"><font color="#0000FF">Pregled grupa</font></p>



<p align="center">Grupa kojoj koncert pripada: 
  <form action = ‘<?= $_SERVER[‘PHP_SELF’]?>’ method = ‘post’>


   <select name="grupa_select" id="grupa_select">
      <?php
    while($row = mysql_fetch_array($sql))
    {
    echo "<option>".$row['grupanaziv']."</option>";
    }
    ?>
    </select>
 </p>

  <p align="center">
    <input type="submit" name="pregled_grupe" id="pregled_grupe" value="- Izaberi grupu-" />
  </p>
  <p>
    <label></label>
  </p>
  </form>
  <?php
   if (isset($_POST['pregled_grupe'])
  $con = mysql_connect($hostname,$db_username,$db_password);
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db($db_name, $con);
        $selektovanje = mysql_query("select id from grupe where grupanaziv ='".$grupa_select."'");
        $grupa_id = mysql_fetch_array($selektovanje);

  $pregled = mysql_query("select * from koncerti where $grupa_id['id'] = grupa_id");

  echo "'$pregled'";

?>

Thanks in advance

php

Edited 10 Years Ago by filipgothic

3 Contributors
26 Replies
3K Views
1 Week Discussion Span
Latest Post 10 Years Ago Latest Post by filipgothic

pritaeas 2,194 ¯\_(ツ)_/¯

10 Years Ago

Missing ) on line 27.

pritaeas 2,194 ¯\_(ツ)_/¯

10 Years Ago

supplied argument is not a valid MySQL result resource

That means an error in your connection or query.

Edited 10 Years Ago by pritaeas

pritaeas 2,194 ¯\_(ツ)_/¯

10 Years Ago

Use mysql_error() to find the problem.

pritaeas 2,194 ¯\_(ツ)_/¯

10 Years Ago

$selektovanje = mysql_query("select id from grupe where grupanaziv ='".$grupa_select."'");

$grupa_select may not have a value. If so then $grupa_id doesn't have one either.

pritaeas 2,194 ¯\_(ツ)_/¯

10 Years Ago

Do you have this somewhere then?

$grupa_select = $_POST['grupa_select'];

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filipgothic 0 Junior Poster in Training · Answer 1 · 2014-06-04T15:35:26+00:00

now new error

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/a9752689/public_html/zadatak2/user.php on line 174

174 is 37

$pregled = mysql_query("select * from koncerti where $grupa_id['id'] = grupa_id");

filipgothic 0 Junior Poster in Training · Answer 2 · 2014-06-04T17:30:32+00:00

I fixed that one, now I have error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 35

which is

$grupa_id = mysql_fetch_array($selektovanje);

filipgothic 0 Junior Poster in Training · Answer 3 · 2014-06-04T20:47:06+00:00

I have similar code in previous selection with same connection and everything and it works, i copied almost everything

 <?php
  if (isset($_POST['unos_koncert'])){
    $enter_koncert = $_POST['enter_koncert'];
    $enter_karte = $_POST['enter_karte'];
    $enter_mesto = $_POST['enter_mesto'];
    $enter_datum = $_POST['enter_datum'];
    $grupa_select = $_POST['grupa_select'];
    if (empty($enter_koncert)){
  die ("ERROR: Niste uneli naziv koncerta!");
  }
  $con = mysql_connect($hostname,$db_username,$db_password);
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db($db_name, $con);
        $sql = mysql_query("select id from grupe where grupanaziv ='".$grupa_select."'");
        $grupa = mysql_fetch_array($sql);
    $sql = "insert into koncerti values (NULL,'".$enter_koncert."','".$enter_karte."','".$enter_mesto."','".$enter_datum."',".$grupa['id'].")";
    if (!mysql_query($sql,$con))
    {
        die('Error: ' . mysql_error());
    }
    echo "Podaci uspesno uneseni u bazu podataka!";
  }
  ?>

$sql = mysql_query("select id from grupe where grupanaziv ='".$grupa_select."'");
$grupa = mysql_fetch_array($sql);

this is same code, just different variables

filipgothic 0 Junior Poster in Training · Answer 4 · 2014-06-04T20:57:12+00:00

alright I fixed this one too now next its soo anoying,

error is

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 35

which is

  $grupa_id = mysql_fetch_array($selektovanje);

filipgothic 0 Junior Poster in Training · Answer 5 · 2014-06-05T09:00:34+00:00

alright I did it and now 3 errors shows

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 172

$grupa_id = mysql_fetch_array($selektovanje);

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/a9752689/public_html/zadatak2/user.php on line 176

if (!mysql_query($con,$selektovanje));

Free Web Hosting
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '= 'grupa_id'' at line 1

filipgothic 0 Junior Poster in Training · Answer 6 · 2014-06-05T09:01:04+00:00

here is whole code that gives bunch of errors

  <?php
   if (isset($_POST['pregled_grupe']))
  $con = mysql_connect($hostname,$db_username,$db_password);
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db($db_name, $con);
        $selektovanje = mysql_query("select id from grupe where grupanaziv =".$grupa_select."");
        $grupa_id = mysql_fetch_array($selektovanje);

  $pregled = mysql_query("select * from koncerti where grupa_id = ".$grupa_id['id']."");

  if (!mysql_query($con,$selektovanje));
    {
    die('Error: ' . mysql_error());
    }

  echo $pregled;

?>

PS I removed errors, but nothing is displayed from base uppon clicking submit
if (isset($_POST['pregled_grupe']))

filipgothic 0 Junior Poster in Training · Answer 7 · 2014-06-05T09:20:37+00:00

I changed code a bit to show rows and after pressing submit button I get this error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 178

which is

while($row = mysql_fetch_array($pregled))

here is code whole code

 <?php
  $con = mysql_connect($hostname,$db_username,$db_password);
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db($db_name, $con);
        $selektovanje = mysql_query("select id from grupe where grupanaziv ='".$grupa_select."'");
        $grupa_id = mysql_fetch_array($selektovanje);

  $pregled = mysql_query("select * from koncerti where grupa_id = ".$grupa_id['id']."");


   if (isset($_POST['pregled_grupe'])) {
        echo "Koncerti za izabranu grupu su: ";
    while($row = mysql_fetch_array($pregled))
  {

    echo "<br>".$row['koncertnaziv'];
    echo "<br>".$row['mesto'];
    echo "<br>".$row['datum'];
    echo "<br>".$row['karte'];
  }
  }



?>

filipgothic 0 Junior Poster in Training · Answer 8 · 2014-06-05T13:58:15+00:00

it has value, because its option to choose band, and I have few bands as option already, and they have data in base, grupanaziv calls all options from base

<p align="center">Grupa kojoj koncert pripada: 
  <form action = "<? $_SERVER[‘PHP_SELF’]?>" method = "post">


   <select name="grupa_select" id="grupa_select">
      <?php
    while($row = mysql_fetch_array($sql))
    {
    echo "<option>".$row['grupanaziv']."</option>";
    }
    ?>
    </select>
 </p>

filipgothic 0 Junior Poster in Training · Answer 9 · 2014-06-05T14:29:41+00:00

no, grupa_select is just for that form, it gets $sql which is previous defined

$sql = mysql_query("select * from grupe");

and it displays grupanaziv as options which is in table grupe, so after I can store ID from that grupa(band) and use it to select data from other table called koncerti (concerts)

guruparthi -6 Newbie Poster · Answer 10 · 2014-06-07T04:51:44+00:00

It means your query contains white space. To avoid this problem you need to add mysql_real_escape_string($val)
ie,
mysql_query("select id from grupe where grupanaziv ='".mysql_real_escape_string($grupa_select)."'");

filipgothic 0 Junior Poster in Training · Answer 11 · 2014-06-07T07:15:24+00:00

I am not geting error there, error stays same with and without that code u gave me

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 178

which is

    while($row = mysql_fetch_array($pregled))

is this ok?

 if (isset($_POST['pregled_grupe'])) {
        echo "Koncerti za izabranu grupu su: ";
    while($row = mysql_fetch_array($pregled))
  {

    echo "<br>".$row['koncertnaziv'];
    echo "<br>".$row['mesto'];
    echo "<br>".$row['datum'];
    echo "<br>".$row['karte'];
  }
  }

what happens if I have more than one record for each koncertnaziv, mesto,datum,karte for selected band(grupanaziv)?

filipgothic 0 Junior Poster in Training · Answer 12 · 2014-06-08T18:26:59+00:00

I am still in need of solution

I will appreciate help

pritaeas 2,194 ¯\_(ツ)_/¯ Moderator Featured Poster · Answer 13 · 2014-06-08T18:39:14+00:00

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 178

If you still have that error, use mysql_error() to find out what the problem is.

filipgothic 0 Junior Poster in Training · Answer 14 · 2014-06-08T19:44:45+00:00

I did, here is whole code

 <?php
   if (isset($_POST['pregled_grupe']))  {
  $con = mysql_connect($hostname,$db_username,$db_password);
    if (!$con)
    {
        die('Could not connect: ' . mysql_error());
    }
    mysql_select_db($db_name, $con);
        $selektovanje = mysql_query("select id from grupe where grupanaziv = '".($grupa_select)."'");
        $grupa_id = mysql_fetch_array($selektovanje);

  $pregled = mysql_query("select * from koncerti where grupa_id = ".$grupa_id['id']."");



        echo "Koncerti za izabranu grupu su: ";
    while($row = mysql_fetch_array($pregled))
  {

    echo "<br>".$row['koncertnaziv'];
    echo "<br>".$row['mesto'];
    echo "<br>".$row['datum'];
    echo "<br>".$row['karte'];
  }
  }


?>

pritaeas 2,194 ¯\_(ツ)_/¯ Moderator Featured Poster · Answer 15 · 2014-06-08T19:46:44+00:00

$selektovanje = mysql_query("select id from grupe where grupanaziv = '".($grupa_select)."'");

$selektovanje = mysql_query("select id from grupe where grupanaziv = '".($grupa_select)."'") or die(mysql_error());
print_r($selektovanje);

Reply with the output.

Am pretty sure $grupa_select doesn't have a value.

filipgothic 0 Junior Poster in Training · Answer 16 · 2014-06-08T19:57:50+00:00

here is code for grupa_select

<p align="center">Grupa kojoj koncert pripada: 
  <form action = "<? $_SERVER[‘PHP_SELF’]?>" method = "post">


   <select name="grupa_select" id="grupa_select">
      <?php
    while($row = mysql_fetch_array($sql))
    {
    echo "<option>".$row['grupanaziv']."</option>";
    }
    ?>
    </select>
 </p>

  <p align="center">
    <input type="submit" name="pregled_grupe" id="pregled_grupe" value="- Izaberi grupu-" />
  </p>
  <p>
    <label></label>
  </p>
  </form>

wait I have it in several forms not just in one, maybe thats problem, let me try change name into something else

pritaeas 2,194 ¯\_(ツ)_/¯ Moderator Featured Poster · Answer 17 · 2014-06-08T19:59:41+00:00

here is code for grupa_select

I know, you already posted that. It was not what I asked for.

I want to know if you add my code above, what you see as output.

filipgothic 0 Junior Poster in Training · Answer 18 · 2014-06-08T20:01:21+00:00

nope still same, error after pressing submit

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a9752689/public_html/zadatak2/user.php on line 181

thats

while($row = mysql_fetch_array($pregled))

filipgothic 0 Junior Poster in Training · Answer 19 · 2014-06-08T20:09:15+00:00

Resource id #11

its 11 just on page load, and after choosing band, its id 12 no matter which band I choose

and I have band with ID 14 only now in base, so idk where he pulls of 11 and 12

filipgothic 0 Junior Poster in Training · Answer 20 · 2014-06-09T14:14:35+00:00

I heard that this mysql commands and version is old, and that might be a problem, I am suggest to use mysqli or pdo

should I change all my code on every php file or try to figure out this line only?

filipgothic 0 Junior Poster in Training · Answer 21 · 2014-06-13T11:04:45+00:00

I am still looking for solution for mysql_fetch_array

if anyone can help I will appreciate