Help with extract() Error should be an Array???

[cybersurfur]

Hi All,

This is my first post on here, since im quite new to php :rolleyes: . Ive been getting this error that i cant for the life of me seem to figour out what it means. :mad: The error shows up as:

Warning: extract() [function.extract]: First argument should be an array in getprod.php on line 34
Line 34 looks like this:

//get our variable passed through the URL
$prodid = $_REQUEST['prodid'];

//get information on the specific product we want
$query = "SELECT * FROM products WHERE product_id='$prodid'";
$results = mysql_query($query)
or die(mysql_error());
$row = mysql_fetch_array($results);
extract($row);
}

where line 'extract($row)' is lne 34

But i thought i had already set it in an array??? Cant see what im doing wrong, can anyone help?? 😕

[bodzan]

There might not be anything returned from your sql query and hence the error...

You might want to do the check first, like:

if(mysql_num_rows($results)>0){//there are results returned
     $row = mysql_fetch_array($results);
     extract($row);
}

[cybersurfur]

bodzan wrote:

There might not be anything returned from your sql query and hence the error...

You might want to do the check first, like:

if(mysql_num_rows($results)>0){//there are results returned
     $row = mysql_fetch_array($results);
     extract($row);
}

wow :eek: thats made the error disappear!!!

But what does that mean??? That the query didnt return a result?? If its not returning a result doesnt that mean theres a problem with it somewhere??

[cybersurfur]

Ive just ran into a similar problem again on a different page, the exact same error:

Warning: extract() [function.extract]: First argument should be an array in cart.php on line 53

but this line is more complicated than the one before:

<?php
$total = 0;
while ($row = mysql_fetch_array($results)) {
echo "<tr>";
extract($row);
$prod = "SELECT * FROM products " .
"WHERE product_id='$carttemp_prodnum'";
$prod2 = mysql_query($prod);
$prod3 = mysql_fetch_array($prod2);
extract($prod3);
echo "<td> ...

line 53 is the one marked in red.
And Ideas??

[bodzan]

...that since your query returned no rows (as in no results) the $row isn't an array...

It might be just the case that your query couldn't return anything because whether your $prodid is empty or not exiting in your database...

The same thing might be true for all other queries where you perform this...

[bong1]

I know this is an old post I am replying back I have the same problem and I tried to follow the suggestions here but it doesn't create an output. It is impossible that I don't have any in the database because I am using the same database like the other search page and it works fine. What am I doing wrong? I hope you can help me solve it as soon as possible. I just want the data to appear in this page calling the selection from other page. Thanks in advance.

<?php 
session_start();

$desname=$_REQUEST['desname'];
if(isset($desname) and strlen ($desname) > 0){
$query = "SELECT * FROM jos_agencyname WHERE desnum='$desname'";
$results = mysql_query($query) or die (mysql_error());
if(mysql_num_rows($results)>0){$row = mysql_fetch_array($results);
extract($row);
}
}

?>
<div align="center" style="background-color:#c6d7ff">

<table width='500px' cellpadding='5px' bgcolor="#c6d7ff">
	<tr>
    <td><?php echo $row[$desnum]; ?></td>
    <td><strong><?php echo $row[$desnum]; ?></strong></td>
	</tr> 
        <tr>
        	<td><form method="post" action="add.php">
            Amount: <input type="text" name="amount" size="8" />
            <input type="hidden" name="desnum" value="<?php echo $desnum ?>" />
            <input type="hidden" name="impact" value="<?php echo $Impact_ID ?>" />
            <input type="submit" name="Submit" value="Add Designation" />
            </form>
            <form method="post" action="ccform.php">
            <input type="submit" name="Submit" value="View Designation" />
            </form>
            </td>
            </tr>
            </table>

[jgetner]

from what i can see you have a problem here

$query = "SELECT * FROM jos_agencyname WHERE desnum='$desname'";

you are tunring the variable $desname into a string

should be

  $query = 'SELECT * FROM  ` jos_agencyname ` WHERE desnum = " $desname" '

also remove the extract function and try to echo the results as such..

 echo $row['desname'] // make sure the row is the right name

[bong1]

HI jgetner, thanks a lot for your fast reply. However, when I tried to use it, it still didn't give anything. When I tried to use print_r, the array is there but there is no data. Is it because I am not getting out from the database? What am I doing wrong? Maybe it helps if I give you the code from my previous page where I need to get the information. I hope you can help me solve this problem. I am stuck just in this page for almost a day now and still can't figure out what is going on. Below is my other page where user can search for the agency and click the hyperlink given to access the whole information from the database.

<?php
$currentPage = $_SERVER["PHP_SELF"];

$maxRows_rs83 = 10;
$pageNum_rs83 = 0;
if (isset($_GET['pageNum_rs83'])) {
  $pageNum_rs83 = $_GET['pageNum_rs83'];
}
$startRow_rs83 = $pageNum_rs83 * $maxRows_rs83;

$colname_rs83 = "-1";
if (isset($_GET['agencyname'])) {
  $colname_rs83 = $_GET['agencyname'];
}
mysql_select_db($databaseconnection, $nameoftable);
$query_rs83 = sprintf("SELECT * FROM jos_agencyname WHERE agencyname LIKE %s", GetSQLValueString("%" . $colname_rs83 . "%", "text"));
$query_limit_rs83 = sprintf("%s LIMIT %d, %d", $query_rs83, $startRow_rs83, $maxRows_rs83);
$rs83 = mysql_query($query_limit_rs83, $Designation) or die(mysql_error());
$row_rs83 = mysql_fetch_assoc($rs83);

if (isset($_GET['totalRows_rs83'])) {
  $totalRows_rs83 = $_GET['totalRows_rs83'];
} else {
  $all_rs83 = mysql_query($query_rs83);
  $totalRows_rs83 = mysql_num_rows($all_rs83);
}
$totalPages_rs83 = ceil($totalRows_rs83/$maxRows_rs83)-1;

$queryString_rs83 = "";
if (!empty($_SERVER['QUERY_STRING'])) {
  $params = explode("&", $_SERVER['QUERY_STRING']);
  $newParams = array();
  foreach ($params as $param) {
    if (stristr($param, "pageNum_rs83") == false && 
        stristr($param, "totalRows_rs83") == false) {
      array_push($newParams, $param);
    }
  }
  if (count($newParams) != 0) {
    $queryString_rs83 = "&" . htmlentities(implode("&", $newParams));
  }
}
$queryString_rs83 = sprintf("&totalRows_rs83=%d%s", $totalRows_rs83, $queryString_rs83);

global $desnum;
global $agencyname;
?>
<form action="getagency.php" method="get" name="frmsearch" style="background-color:#c6d7ff">
<table width="100%" border="0" cellpadding="2" bgcolor="#ffffff" id="search">
  <tr>
    <td width="24%">Search Agency</td>
    <td width="61%"><input name="agencyname" type="text" style="width:250" /></td>
    <td width="15%"><input name="Submit" type="submit" value="Search" /></td>
  </tr>
</table>



<?php do { ?>
  <table width="100%" border="0" cellpadding="2">
    <tr>
      <td width="16%"><?php echo "<a href='getagency.php?desname=" .$desnum ."'>";?><?php echo $row_rs83['desnum']; ?><?php echo "</a>"; ?> </td>
      <td width="79%"><?php echo "<a href='getagency.php?desname=" .$agencyname ."'>";?> <?php echo $row_rs83['agencyname']; ?><?php echo "</a>"; ?></td>


</tr>
  </table>
  <?php } while ($row_rs83 = mysql_fetch_assoc($rs83)); ?>
  <table width="100%" border="0" cellpadding="2">
    <tr>
      <td width="16%"><?php echo "<a href='getagency.php?desname=".$desnum ."'>";?><?php echo $_SESSION[$row_rs83['desnum']]; ?><?php echo "</a>"; ?> </td>
      <td width="79%"><?php echo "<a href='getagency.php?desname=".$agencyname ."'>";?> <?php echo $_SESSION[$row_rs83['agencyname']]; ?><?php echo "</a>"; ?></td>

</tr>
  </table>





</form>
<?php
mysql_free_result($rs83);
?>

[jgetner]

when you go into the database through phpmyadmin does your tables have any results?

[bong1]

thank you so much for your help in this. Yes, when I tried to do a search I have an output. It just that it is not grabbing what information I want. Hope you can help me figure it out. Thanks.

[johanafm]

jgetner;10965697 wrote:

from what i can see you have a problem here

$query = "SELECT * FROM jos_agencyname WHERE desnum='$desname'";

you are tunring the variable $desname into a string

should be

  $query = 'SELECT * FROM  ` jos_agencyname ` WHERE desnum = " $desname" '

Not at all. The first version is syntactically correct, both from a PHP and a MySql perspective. $desname allready is a string (most likely), and in a weakly types language it doesn't matter much anyway. Since the string literal is delimited by " (rather than ') string interpolation kicks in and $desname is replaced by whatever the variable contains.
Assuming $desname contains 'John', the string will become

$query = "SELECT * FROM jos_agencyname WHERE desnum='John'"

The single quote is used as delimiter for the string John, which is as it should be. The (non standard SQL, standard uses ") backtick is only needed to escape identifiers (such as fields or tables) if if the identifier is also a reserved word such as (case-insensitive) "date" or "int". And if you do use to delimit the identifier, EVERYTHING inside it is interpreted as belonging to the identifier. As such, there is a difference between these[code=php]
$table = "jos_agencyname"; # this matches the table
$table = " jos_agencyname `"; # this doesn't
[/code]

I find the issue much more likely to be a programming logic error, since the field identifier in the where clause is "desnum" while the variable used is called "$desname".

[bong1]

Hi, thanks again for your help. If it is a programming error would it be possible to tell me where? Basically, I wanted to get the selected agency via the input so I can get the information from database. I followed the book but it doesn't get the right output. Would it be possible to help me? Thanks in advance.

[bong1]

Oh, by the way, I guess it is important also to know that I am using joomla for this. Thanks. Any insights is very much appreciated to get this done.