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PHP - Automatic Creation Of List When Getting Text From Db
Hi all,
Weird one: I am re-working a site for a client. Their old site currently grabs text from a DB and somehow automatically puts it into a <ul>. The only thing to signify a <li> is a linebreak. So if I entered: Item 1 Item 2 Item 3 into the database, they would come out looking like: Item1 Item 2 Item 3 on the site. Anyone know of a way to recreate this? An example of this in action is he http://www.postureperfection.com.au/Chairs/Amore_Chair/p/495/ (please remember - I am redesigning the site!) Thanks heaps! Similar TutorialsHi, In my mysql database i have a text input option, in the registration form and edit my details form i have a multiple select dropdown list, which user selects options to populate the text input box, which ultimately populates the text field in the mysql database. All works perfectly. The dropdownlist consists of 3 parts <optgroups> first is current selection (what is the usesr current selection)works fine, The second <optgroup> is existing words, what words we(the site) have given as options, and the third <optgroup> is the words that others have used. This is where im having a small problem. Because its a text field when i call the data from the database, it calls the entire text box as a single option in my select list.. I want to break the words in the text field (at the comma) and have them listed each one as an option in the select list. Example what i need: Words in text box:(my input allows the "comma") word1, word2, word3, word4, word5, word6, How i want them called/displayed: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> here's my code: $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $original_functionsexperience =doSelectSql($query,1); $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $functionsexperiencelist=doSelectSql($query); $funcexpList ="<select multiple=\"multiple\" onchange=\"setFunctionsexperience(this.options)\">"; foreach ($functionsexperiencelist as $functionsexperienceal) { $selected=""; if ($functionsexperienceals->allwords == $original_functionsexperience) $selected=' selected="selected"'; $allwords=$functionsexperienceal->allwords; $funcexpList .= "<optgroup label=\"Current selection\"> <option value=\"".$allwords."\" ".$selected." >".$allwords."</option> </optgroup> <optgroup label=\"Existing Words\"> <option value=\"existing1,\">existing1</option> <option value=\"existing2,\">existing2</option> <option value=\"existing3,\">existing3</option> <option value=\"existing4,\">existing4</option> <option value=\"existing5,\">existing5</option> <option value=\"existing6,\">existing6</option> </optgroup> <optgroup label=\"Others added\"> //heres problem <option value=\"".$allwordsgeneral."\">".$allwordsgeneral."</option> </optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; The result im getting for optgroup others added: word1, word2, word3, word4, word5, how can i get it like this: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> Hello all , here is another problem of my project. I need to create a textarea , drop down list and submit button . At first , I can type whatever I want in the textarea , but for certain part I can just choose the word I want from drop down list and click submit , then the word will appear in the textarea as my next word . But I have no idea how to make this works , is there any simple example for this function ? Thanks for any help provided . (If this is a regex question and should be moved, I apologize.. wasn't sure since I don't know the answer) If my $description is the following, how would I strip away all the stuff before the list and after it and keep only the contents of the ul?: random text.. blah blah blah.. more text.. blah blah list item 1 list item 2 list item 3 random text.. blah blah blah.. more text.. blah blah so far I have:
$crawler->filter('li.date, dt > a, li.style')->each(function ($node) {
$output = $node->text()."\n";
$array = explode("\n", $output);
$data = array_chunk($array, 3);
// print_r($data);
foreach ($data as $row) {
[$date, $title, $type] = array_pad($row, 3, null);
// here you have your variables
print 'Date: '.$date;
print 'Title: '.$title;
print 'Type: '.$type;
}
});
and this outputs: QuoteDate: Saturday 13th Jun 2020Title: Type: Date: Alice in Chains UKTitle: Type: Date: Metal/ RockTitle: Type: Date: Monday 29th Jun 2020Title: Type: Date: Cage the ElephantTitle: Type: Date: Funk/ RockTitle: Type: Date: Friday 31st Jul 2020Title: Type: Date: The CureheadsTitle: Type: Date: Indie/ RockTitle: Type: Date: Thursday 20th Aug 2020Title: Type: Date: CreeperTitle: Type: Date: PunkTitle: Type: Date: Saturday 22nd Aug 2020Title: Type: Date: Fleetwood BacTitle: Type: Date: Pop/ RockTitle: Type: Date: Monday 31st Aug 2020Title: Type: Date: GZATitle: Type: Date: Hip HopTitle: Type: Date: Saturday 5th Sep 2020Title: Type: Date: Cock SparrerTitle: Type: Date: Only $date variable is set for each?! please help??
it should be: Date: Saturday 13th Jun 2020Title: Alice in Chains UKType: Metal/ Rock Date: Monday 29th Jun 2020Title: Cage the ElephantType: Funk/ Rock and so on.... I have a form where I have inserted 7 pre-populated relational lists. All of the information is pulling correctly from the databases, but when it posts, it's posting the value "ids" instead of the chosen text. The files a www.kcwell.com/gcc_form.php and www.kcwell.com/gccsuccess_form.php How do I set up a query to obtain the data that I need? Help! Hi guys, basically here pull out the data from database then creating taxt field automatically and submit into anther table. everything works fine but data not inserting in to the table. could you guys check my code please? <?php $con = mysql_connect("localhost","root",""); mysql_select_db("uni", $con)or trigger_error('MySQL error: ' . mysql_error()); ?> <?php $result = mysql_query("SELECT * FROM course") or trigger_error('MySQL error: ' . mysql_error()); echo '<select name ="cid[]">'; while($row = mysql_fetch_array($result)) { echo '<option value="' . $row['CourseID'] . '">'. $row['CourseID'] .'</option>'; } echo '</select>'; //------------------ ?> <?php if(!empty($_POST["submit"])) { $value = empty($_POST['question']) ? 0 : $_POST['question']; ?> <form name="form1" method="post" action="result.php"> <?php for($i=0;$i<$value;$i++) { echo 'Question NO: <input type="text" name="qno[]" size="2" maxlength="2" class="style10"> Enter Marks: <input type="text" name="marks[]" size="3" maxlength="3" class="style10"><br>'; } ?> <label> <br /> <br /> <input type="submit" name="Submit" value="Submit" class="style10"> </label> </form> <?php } else{ ?> <form method="POST" action="#"> <label> <span class="style10">Enter the Number of Question</span> <input name="question" type="text" class="style10" size="2" maxlength="2"> </label> <input name="submit" type="submit" class="style10" value="Submit"> </form> <?php }?> result.php <?php $con = mysql_connect("localhost","root","") or die('Could not connect: ' . mysql_error()); mysql_select_db("uni",$con) or die('Could not connect: ' . mysql_error()); foreach ($_POST['cid'] as $c) {$cid [] = $c;} foreach($_POST['qno'] as $q){$qno[] = $q;} foreach($_POST['marks'] as $m){$marks[] = $m;} $ct = 0; for($i=0;$i<count($qno);$i++) { $sql="INSERT INTO examquesion (CourseID,QuesionNo,MarksAllocated) VALUES('$cid[$i]','$qno[$i]','$marks[$i]')"; mysql_query($sql,$con) or die('Error: ' . mysql_error()); $ct++; } echo "$ct record(s) added"; mysql_close($con) ?> Hi, I'm new to PHP/MySQL and need some help getting my query to work for my selection list: The selection list is built with: <form action='processformmissing.php' method='POST'> <fieldset> <legend>Choose Department</legend> <select name='depart'> <option value=''></option> <?php while ($row = mysqli_fetch_array($result)) { extract($row); echo "<option value='$department'>$department</option>\n"; } ?> </select> <p><input type='submit' value='Select Department' /></p> </fieldset> </form> The data is then sent to: $depart = $_POST['depart']; $deptlike = "%".$depart."%"; echo "<p>$depart</p>"; echo "<p>$deptlike</p>"; $query = "SELECT * FROM lifecerts INNER JOIN employees ON lifecerts.cid = employees.cid WHERE department LIKE '$deptlike' ORDER BY employees.name"; Hitting the submit button from my selection list form seems to be working fine because when I echo my data ($depart and $deptlike) it is giving me the correct value, but the query doesn't give me any results. However, if my post data comes from a text box instead of a selection list, my query works fine. Any thoughts on what I'm doing wrong??? Many thanks! With the following code i can create an xml file. Code: [Select] <?php $myXML = new SimpleXMLElement("<myroot></myroot>"); $title= $myXML->addChild('title'); $title->addAttribute('number','12'); $titleName= $title->addChild('titleName', 'title1'); $titleLink= $title->addChild('titleLink', 'link1'); Header('Content-type: text/xml'); echo $myXML->asXML(); ?> But when i check the validity of xml file http://www.validome.org/xml/validate/ This error occurs: Can not find declaration of element 'myroot'. I suppose that the problem is missing of !DOCTYPE and !ELEMENT lines. How can i create valid XML documents with PHP automatically?? Is it possible to make it without writing doctype and element types for the whole element types of xml by hand $title, $titleName and $titleLink Thank you i have mysql community server 5.5 workbench 5.2 php 5.3.8 apache http server 2.2 summary: can i write the same code using less variables for the file name and content? okay so i managed to write a code that creates a filename, and each new filename is 1 higher. colony1 colony2 colony3. it also uses the same mysql query created variable to write idcol as 1 higher in the new file than in the previous written file. however to write the code i had to use a buttload of variables. here is the code, the resultant filename and the resultant content of the created file. is there a way to write such a code with less than 15 variables? the problem seems to be that when writing text to a file, i cannot insert certain characters outside of a variable into the final variable which is put in the $string variable input into the fwrite() function. File Name: Colony$.php where $=the last entry in the idcol column in table coltest. File Code: Code: [Select] <?php $idcol = $; include('Colony0.php'); ?> </body> </html> where $=the last entry in the idcol column in table coltest. Code: Code: [Select] <?php $dbhost = 'localhost:3306'; $dbuser = 'root'; $dbpass = 'root'; $dbname = 'aosdb'; $conn = mysql_connect($dbhost,$dbuser,$dbpass) or die ('Error connecting to mysql'); mysql_select_db($dbname); $query = "SELECT idcol FROM coltest ORDER BY idcol DESC LIMIT 1"; $result = mysql_query($query); $id = mysql_result($result, 0); $vcol = "Colony"; $vcolp = ".php"; $File = "$vcol$id$vcolp"; $FileHandle = fopen($File, 'w') or die("can't open file"); fclose($FileHandle); $File = "$vcol$id$vcolp"; $myFile = $File; $fh = fopen($myFile, 'w') or die("can't open file"); $idw = "$"; $icol= "idcol"; $ique = "?"; $iphp = "Colony0.php"; $ief = "<"; $ieb = ">"; $irese = "= "; $iresn = ";"; $iii= "include('$iphp');"; $ihtml ="</body> </html>"; $hph = "php"; $string = "$ief$ique$hph $idw$icol $irese$id$iresn $iii $ique$ieb $ihtml"; fwrite($fh, $string); fclose($fh); ?> I am attempting to create an image and I have run into a snag. I want to add both text and another image to the image I am making. Here is the code so far minus unnecessary parts: header ("Content-type: image/png"); $name = $fetch_user['username']; $rank = $fetch_rank['rank_title']; $img_url = 'path''. $fetch_rank['rank_image']; $im = @imagecreatefrompng("$img_url") or die("Cannot Initialize new GD image stream"); // try changing this as well $font = 4; $width = imagefontwidth($font) * strlen($string) ; $height = imagefontheight($font) ; $im = imagecreatefrompng("image.png"); $x = imagesx($im) - $width ; $y = imagesy($im) - $height; $backgroundColor = imagecolorallocate ($im, 255, 255, 255); $textColor = imagecolorallocate ($im, 255, 255, 255); //imagestring ($im, $font, $x, $y, $string, $textColor); imagestring ($im, $font, 100, 10, $name, $textColor); imagestring ($im, $font, 100, 22, $rank, $textColor); imagepng($im); Everything works UNTIL I added in this part: $im = @imagecreatefrompng("$img_url") or die("Cannot Initialize new GD image stream"); Any idea what I am doing wrong? All that comes up is "Cannot Initialize new GD image stream"; and I know the URL's work. Whats is the best way to create a forum using php? Should I use any frameworks? If not is there any guide/tutorial/book that descibe this procedure? Thank you. Hi, I am wanting to check to see if a folder exists, if not create it. PHP on a windows machine so I would like to know how to handle the folder separators. I am wanting to use a path which includes the drive letter, such as... $foldername ="C:\folder1\images\$checkfolder" I was thinking about using something like this: if(!is_dir($foldername)) mkdir ("$foldername",777); What would be the best way of tackling this using windows paths? I am trying to create a basic image with GD library, but my browser says that the image cannot be displayed because it contains errors... I have no idea why. Here's my code: Code: [Select] switch($_GET['case']){ case "progressbar": header('Content-Type: image/jpeg'); $barWidth = 6; $barHeight = 14; $barPadding = 2; $imageWidth = ($barWidth * 10) + (10 * $barPadding); $imageHeight = ($barHeight + ($barPadding * 2)); $img = @imagecreate($imageWidth, $imageHeight); $imgBorder = imagecolorallocate($img, 0, 0, 0); //imagefilledrectangle($img, 1, $imageHeight -1, $imageWidth - 1, $imageHeight - 1, $imgBorder); $output = "Hello, world!"; imagestring($img, 1, 4, 4, $output, $imgBorder); imagejpeg($img); imagedestroy($img); break; } I have a file called image.php that i use to create images and I use the URL image.php?case=progressbar. I am not sure what to look for to even get started. If some one could help point me in the right direction that would be appreciated.
I created a customer system that tracks contracts and invoices ect. Where do I start if I would want a customer to be able to go to the website, sign up, and their own version of the system would be created.
Do I just write a script that creates a new database with the correct credentials? Do all of the users access the same files but just have different database. Do I have to copy the directory of files to a specific folder for each customer?
I know there must be tons of websites that once you sign up for example a calendar app that keeps your data seperate from everyone else.
Any help is appreciated.
Thanks
Jack
Day
I'm using Codeigniter and inside the framework I create a file with PHP code If I try to point to it, the web-server doesn't deliver due to permission issues.
All this is done in a 'register function' where I create some fields in a SQL
If I put the user credentials into the database 'by hand' and likewise when I
I'm pretty sure this issue has been existed before and I'd appreciate any help.
Thanks for your help in forward. Gee Edited August 17, 2020 by bogusHey, guys, in the following code, why i cannot see my ~/test.txt file? It seem he is not doing what i want, to create a .txt file with 'is just a test'. I'm using Debian.
<?php
$f = fopen("~/test.txt","w+");
fwrite($f,'is just a test');
fclose($f);
?>
Thank you!
Hi All I am trying to insert a feed on my site that automatically updates the football league etc like this one http://news.bbc.co.uk/sport1/hi/football/eng_div_1/table/default.stm is there a way to do this please? thanks This topic has been moved to mod_rewrite. http://www.phpfreaks.com/forums/index.php?topic=321416.0 Hi So I have successfully set up a website and database where a user can create a listing and view listings on a listing details page. The viewing can only be done, however, only from either from either running a search query from the search form or from clicking one of the listings from the listing table. The listings are not being indexed from the search engines from the direct page, instead from the listing table which is about 3,000 listings, not very user friendly to land on this page and have to go through the table or use the table filter. Here's what I am trying to accomplish: A web page that has a dynamic presence / URL like a Wordpress page: http://www.mysite.com/listing_title/ The title of the page has the listing information in it: This page is about a $listing_title $listing_item listing number $listing_number that is online! Each individual Listing / page is indexed on the search engines and has a direct link to it I am familiar with PHP and have created this site but I am far from an expert! Please give me a code example or link to one because if you just say "do this to this and that" you will totally loose me. Thank you so much for any help. I was not sure on which board to put this thread, so please move it if there is a more appropriate board. I am wanting to create a website that displays images which I have put into a database. I would like the number of images on one page not to exceed 10, for example, and when the number of images in the database exceeds this number, a new page will be created. I would like it to function similar to Google Images where the number of pages is dictated by the number of images in the database. I cant figure out how to achieve this, so any pointers or thoughts would be great Thanks, Matlab |
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