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ucwords(strtolower(?)) and supplied argument is not a valid ?


plodos

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Problem 1 )

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/wasetorg/public_html/search.php on line 177

$data = mysql_query("select * from person where $type like '%$search%' ORDER BY lname ASC LIMIT 0,30"); 
                     
while($info=mysql_fetch_array($data))
{
	if($info['no']=="1")
	{
	echo"			{ucwords(strtolower($info['lname']))} {ucwords(strtolower($info['fname']))} ";  //line 177
	}
}

 

Problem 2)

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:.........

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:.........

 

while($info=mysql_fetch_array($data))
{
	if($info['no']=="1")
	{
	echo"			{$info['lname']} {$info['fname']}";
	}

 

Some of the mysql datas are uppercase/lowercase like ERIK, Erik, erik....maybe because of that it gives "supplied argument is not a valid" worning...

 

where $type like '%$search%'

Im using like operators, if I convert all data with ucwords(), I can match the true datas....

 

How can I solve these problems? I didnt do it....

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	echo"			{ucwords(strtolower($info['lname']))} {ucwords(strtolower($info['fname']))} ";  //line 177

 

Two things with that:

- you can't call functions within "" quotes.

- if you use array variable within "" quotes you must surround it with {} braces. <- this one is actually causing a warning

 

You should do:

	echo "{".ucwords(strtolower($info['lname']))."} {".ucwords(strtolower($info['fname']))."}";  //line 177

 

That's assuming you want {} around last name and first name (I think you might put them by mistake)

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echo ucwords(strtolower($info['lname']))." ".ucwords(strtolower($info['fname']))."}, {".$info['country']." (Nat) <br>;
echo ucwords(strtolower($info['university'])) <br>;

 

I didnt apply the code

 

Parse error: syntax error, unexpected '(', expecting ',' or ';

 

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echo ucwords(strtolower($info['lname']))." ".ucwords(strtolower($info['fname']))."}, {".$info['country']." (Nat) <br>";
echo ucwords(strtolower($info['university']))." <br>";

 

 

Try to find an editor that would highlight your code. It will make spotting mistakes like this much easier

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$Rows = mysql_num_rows(mysql_query("select * from person where $type like '%$search%'")) or die (mysql_error ()); 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/public_html/search.php on line 187

-----------------------and------------------

$data = mysql_query("select * from person where $type like '%$search%' ORDER BY lname ASC LIMIT $start,$limit") or die (mysql_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 'like '%%'' at line 1

 

What are these problems?

 

 

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Make this:

$Rows = mysql_num_rows(mysql_query("select * from person where $type like '%$search%'")) or die (mysql_error ()); 

 

This:

$query = mysql_query("select * from person where $type like '%$search%'") or die(mysql_error());
$Rows = mysql_num_rows($query);

Basically that meant that the query is most likely failing and returning a MySQL Resource ID.

 

The second error is saying that $search is null, so make sure your variables are being set properly.

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					<?php
include("dbconfig.php");

function title_case($title) {
  // Our array of 'small words' which shouldn't be capitalised if
  // they aren't the first word.  Add your own words to taste.
  $smallwordsarray = array(
    'of','a','the','and','an','or','nor','but','is','if','then','else','when',
    'at','from','by','on','off','for','in','out','over','to','into','with'
    );
  // Split the string into separate words
  $words = explode(' ', $title);
  foreach ($words as $key => $word)
  {
    // If this word is the first, or it's not one of our small words, capitalise it
    // with ucwords().
    if ($key == 0 or !in_array($word, $smallwordsarray))
      $words[$key] = ucwords(strtolower($word));
  }
  // Join the words back into a string
  $newtitle = implode(' ', $words);
  return $newtitle;
}


$search = mysql_real_escape_string($_POST['search']);
$type = mysql_real_escape_string($_POST['type']);

$page = $_GET["page"]; 
if(empty($page) or !is_numeric($page)){ 
$page = 1;
}

$limit = 25; 

$query1 = mysql_query("select * from person where $type like '%$search%'") or die (mysql_error () );
$rowNumber = mysql_num_rows($query1); 

$pageNumber = ceil($rowNumber / $limit); 
$start = ($page-1)*$limit; 

$data = mysql_query("select * from person where $type like '%$search%' ORDER BY lname ASC LIMIT $start,$limit") or die (mysql_error ()); 
                     
while($info=mysql_fetch_array($data))
{
	if($info['committee_no']=="1")
	{
                     echo title_case($info['lname'])." ".title_case($info['fname']).", ".$info['country']." <br>";
	}
	if($info['no']=="2")
	{
                     echo title_case($info['lname'])." ".title_case($info['fname']).", ".$info['country']."  <br>";
	}
}


if($page > 1){ 
$back = $page-1;
echo "<p align=\"center\"><b><a href=\"search.php?page=$back\"><< Back </a>";

}else{ 

echo "<p align=\"center\"><b> << Back ";
}

echo " | ";

if($page>=$pageNumber){ 

echo "Next >> </b></p>";

}else{ 

$next = $page+1;

echo "<a href=\"search.php?page=$next\">Next >></a></b></p>";

}
?>

 

Im trying to using paging 0-25, 25-50 .......

 

first page is listing all datas 0-25, but when I click to next page....program is giving there errors?...

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 'like '%%'' at line 1

 

I used this paging code before, i didnt face any prbblem....but now this scipts first page is showing the datas 0-25, why the second page is giving this error ????

 

Some of the mysql datas are uppercase/lowercase like ERIK, Erik, erik...and im trying the search with lowercase...it can be an error?

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$_POST['search'] and $_POST['type'] is not blank....

for example im searching 'Mobile', program finds 43 rows....

 

First page is listing 25 records, but when I click the next page, just an error....next page must show 18 records...

 

 

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Because your $search and $type variables are not set when you go from page to page.

 

You're using links to switch between pages. Not only links are $_GET not $_POST, but your links don't even conatin these variables.

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