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ger_mac74

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  1. yes desleted connect details for security reasons. Have tried echoing parts and everything seems fine. i am newish to php. The script is running fine except that when the page loads up first the html stuff below the php file is not appearing on the website. but when the search is done the search finds the articles and the html stuff after php script appears fine then. As for the second part for problem it is saying that 25 reults are found but the link to next 10>> is only displaying the same page with first 10 results. i found the script online. cheers
  2. Hi Here is the form part of html and the php script that follows after it. The html file is called search1.php which is also what the action is calling. Also having prob displaying results. First 10 results are being displayed fine but the next 10>> link is not working..same page being return. Thanks for coming back to me. Cheers [code]<p> <br></br> <form name="form" action="search1.php" method="get">   <input type="text" name="q" />   <input type="submit" name="Submit" value="Search" /> </form> <?php   // Get the search variable from URL   $var = @$_GET['q'];   $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit =10; // check for an empty string and display a message. if ($trimmed == "")   {   echo "<p>Please enter a search...</p>";   exit;   } // check for a search parameter if (!isset($var))   {   echo "<p>We dont seem to have a search parameter!</p>";   exit;   } //connect to database mysql_connect("","",""); //specify database   **** mysql_select_db("") or die("Unable to select database"); // Build SQL Query $query = "SELECT * FROM articles1 WHERE (description LIKE '%$trimmed%' ) OR (articlename LIKE '%$trimmed%' ) ORDER BY description"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0)   {   echo "Results";   echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>"; // google echo "<p><a href=\"http://www.google.com/search?q="   . $trimmed . "\" target=\"_blank\" title=\"Look up   " . $trimmed . " on Google\">Click here</a> to try the   search on google</p>";   } else{ // next determine if s has been passed to script, if not use 0   if (empty($s)) {   $s=0;   } // get results   $query .= " limit $s,$limit";   $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: &quot;" . $var . "&quot;</p>"; // begin to show results set echo "Results: <br>"; $count = 1 + $s; // now you can display the results returned   while ($row= mysql_fetch_array($result)) {            $title = $row["articlename"];            $arturl  = $row["articleurl"];            echo "$count.) <a href='$arturl'>$title</a><br>";            $count++; } $currPage = (($s/$limit) + 1); //break before paging   echo "<br />";   // next we need to do the links to other results   if ($s>=1) { // bypass PREV link if s is 0   $prevs=($s-$limit);   print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt;   Prev 10</a>&nbsp&nbsp;";   } // calculate number of pages needing links   $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division   if ($numrows%$limit) {   // has remainder so add one page   $pages++;   } // check to see if last page   if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {   // not last page so give NEXT link   $news=$s+$limit;   echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";   } $a = $s + ($limit);   if ($a > $numrows) { $a = $numrows; }   $b = $s + 1;   echo "<p>Showing results $b to $a of $numrows</p>"; } ?>[/code][s]
  3. Hi all I am running a php search script within a html webpage. The top part of the web page appears ok when viewed, as does the script part ..but the part of html code after the script is not appearing on page until the script is called. The script is calling itself..i think this may be where the problem is coming from?? I would like to keep the script in the same file. Any ides whats happenin?
  4. does anyone have any ideas on how i would fix one or two incorrect lines in this script? thanks
  5. does anyone have any ideas how i could fix one or two incorrect lines in this script??
  6. Hi there I am using this search script that I found online, in a web page that I am designing. It is searching a table in a database. When I open the file the search script finds and lists the first 10 results but the link to Next 10>> is not working. It does find enough results to have a next page as it is also printing "Showing results 1 to 10 of 35"" below the link. I am new to php and am having problems here. Can anyone help me out? Please! Thank you... Geraldine [code]<form name="form" action="search.php" method="get">   <input type="text" name="q" />   <input type="submit" name="Submit" value="Search" /> </form> <?php   // Get the search variable from URL   $var = @$_GET['q'];   $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit =10; // check for an empty string and display a message. if ($trimmed == "")   {   echo "<p>Please enter a search...</p>";   exit;   } // check for a search parameter if (!isset($var))   {   echo "<p>We dont seem to have a search parameter!</p>";   exit;   } //connect to database mysql_connect("cosmos.ucc.ie","gpmc1","itahkauf"); //specify database   **** mysql_select_db("2006_gpmc1") or die("Unable to select database"); // Build SQL Query $query = "SELECT * FROM articles1 WHERE (description LIKE '%$trimmed%' ) OR (articlename LIKE '%$trimmed%' ) ORDER BY description"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0)   {   echo "Results";   echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; // google echo "<p><a href=\"http://www.google.com/search?q="   . $trimmed . "\" target=\"_blank\" title=\"Look up   " . $trimmed . " on Google\">Click here</a> to try the   search on google</p>";   } // next determine if s has been passed to script, if not use 0   if (empty($s)) {   $s=0;   } // get results   $query .= " limit $s,$limit";   $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results: <br>"; $count = 1 + $s; // now you can display the results returned   while ($row= mysql_fetch_array($result)) {            $title = $row["articlename"];            $arturl  = $row["articleurl"];            echo "$count.) <a href='$arturl'>$title</a><br>";            $count++; } $currPage = (($s/$limit) + 1); //break before paging   echo "<br />";   // next we need to do the links to other results   if ($s>=1) { // bypass PREV link if s is 0   $prevs=($s-$limit);   print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><<   Prev 10</a>  ";   } // calculate number of pages needing links   $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division   if ($numrows%$limit) {   // has remainder so add one page   $pages++;   } // check to see if last page   if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {   // not last page so give NEXT link   $news=$s+$limit;   echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>";   } $a = $s + ($limit);   if ($a > $numrows) { $a = $numrows; }   $b = $s + 1;   echo "<p>Showing results $b to $a of $numrows</p>"; ?>[/code]
  7. Hi again I know the code is a bit long but my guess is that there is not much wrong with it. I found the code online and am new to php myself. Mable somebody experienced could help me out...please... thank you
  8. Hi there does anyone have any ideas on how to fix this. The search script is essentially working but i am unable to view any links after the 10th one found.....the next 10>> link is not working/displaying results...when clicked on nothing happens. Cheers
  9. Hi there I am using this search script in a web page that I am designing. It is searching a table in a database. When I open the file the search script finds and lists the first 10 results but the link to Next 10>> is not working. It does find enough results to have a next page as it is also printing "Showing results 1 to 10 of 35"" below the link. I am new enough to php and am having problems here. Can anyone help me out? Cheers [code] <form name="form" action="search.php" method="get">   <input type="text" name="q" />   <input type="submit" name="Submit" value="Search" /> </form> <?php   // Get the search variable from URL   $var = @$_GET['q'];   $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit =10; // check for an empty string and display a message. if ($trimmed == "")   {   echo "<p>Please enter a search...</p>";   exit;   } // check for a search parameter if (!isset($var))   {   echo "<p>We dont seem to have a search parameter!</p>";   exit;   } //connect to database mysql_connect("cosmos.ucc.ie","gpmc1","itahkauf"); //specify database   **** mysql_select_db("2006_gpmc1") or die("Unable to select database"); // Build SQL Query $query = "SELECT * FROM articles1 WHERE (description LIKE '%$trimmed%' ) OR (articlename LIKE '%$trimmed%' ) ORDER BY description"; $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0)   {   echo "Results";   echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; // google echo "<p><a href=\"http://www.google.com/search?q="   . $trimmed . "\" target=\"_blank\" title=\"Look up   " . $trimmed . " on Google\">Click here</a> to try the   search on google</p>";   } // next determine if s has been passed to script, if not use 0   if (empty($s)) {   $s=0;   } // get results   $query .= " limit $s,$limit";   $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results: <br>"; $count = 1 + $s; // now you can display the results returned   while ($row= mysql_fetch_array($result)) {            $title = $row["articlename"];            $arturl  = $row["articleurl"];            echo "$count.) <a href='$arturl'>$title</a><br>";            $count++; } $currPage = (($s/$limit) + 1); //break before paging   echo "<br />";   // next we need to do the links to other results   if ($s>=1) { // bypass PREV link if s is 0   $prevs=($s-$limit);   print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><<   Prev 10</a> ";   } // calculate number of pages needing links   $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division   if ($numrows%$limit) {   // has remainder so add one page   $pages++;   } // check to see if last page   if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {   // not last page so give NEXT link   $news=$s+$limit;   echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>";   } $a = $s + ($limit);   if ($a > $numrows) { $a = $numrows; }   $b = $s + 1;   echo "<p>Showing results $b to $a of $numrows</p>"; ?>[/code] [b]EDITED BY OBER: PLEASE USE CODE TAGS WHEN POSTING CODE.[/b]
  10. Thanks for that. I looked at the manual and cant find what to do. I have tried putting in a letter instead of " " and the letter prints out fine which means the loop and count are working ok. It just doesnt seem to want to print an empty space. I also tried echoing "\s" and "&nbsp;" but neither work. Is there another system I should use to actually print a space or do I use a differnt function. Thanks again.
  11. That's great. Working fine now. I had just left out a comma after articlename in my select query. When it is printing out now the article name is beside the number listed. This is what I want but when it gets to number 10.) it is one space further out the page compared to 9.) Is there a piece of code that I could fit in here to fix this and make 1 to 9 appear one space further out? Not too worried about what happens when it gets to 100. Looks like this at the moment: 8.) Text...... 9.) Text...... 10.)Text...... 11.)Text...... 12.)Text...... Not as obvious here but after 9 text appears out one space further because of extra digit. Hope you can help me out. Regards...
  12. Hi there I am working on a php script for a webpage that will grab two columns from a database (articlename and articleurl) and then number and display just the article name column. This article name column will hopefully appear as a link and when clicked on, the page will then divert to the articles url. Also I am new enough to php and programming so be easy on me! Here is the code that I have so far: <?php $db = mysql_connect("----", "----", "----"); mysql_select_db("-------", $db); $query = "SELECT articlename articleurl FROM articles"; $result = mysql_query($query) or die("Couldn't execute query"); ?> <?php echo "Current List Of Articles: <br>"; $count = 1; while ($row= mysql_fetch_array($result)){ $artname = $row["articlename"]; $arturl = $row["articleurl"]; echo "$count.) <a href='$arturl'>$artname</a><br>"; $count++; } ?> Can anyone see what I am doing wrong? At the moment the code seems to be counting the number of entries in the database but no articlename links seem to be appearing!! Thanks in advance!!!
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