headmine

Exactly what I needed. I knew about spaces and a str_replace is exactly what I was going to do. Thanks so much for your help.

I have a string of Data. Inside this string there is a line of text that says Where: 1234 Main Street. What I would like to do is grab the address that comes after "Where:" and have the output with a link to google maps ex Where: <a href="http://maps.google.com?q=1234+Main+Street">1234 Main Street</a> Would I use preg_replace for this? Or am I attacking this the wrong way?

Ok. I got it. So here it is for anyone else that might run into this problem $sql = "SELECT value2, COUNT(value2) as total FROM mydb WHERE value1 = 'state' GROUP BY value2" foreach ($sql as $state) { echo $sql->meta_value . ' (' . $sql->total .')<br />'; } This would echo: state (count) CA (3323) NV (93) etc.

OK. I've gotten it this far. SELECT value1, COUNT(value2) FROM mydb WHERE value1 = 'state' GROUP BY value1" foreach ($v as $vn) { echo $vn->value1 . '<br />'; } This prints out each state listed . How do I get the numbers next to it? Any help would be appreciated!

OK! Breakthru! I was able to do this within the SQL database. SELECT value2, COUNT(value1) As total FROM mydb WHERE value1 = 'state' GROUP BY value2 When I do this in SQL it works beautifully! It breaks down the states and how many records there are for each. When I do this in php. I only get the first Value! =( How do I get it to display all the values? THANKS!

I have a database with 4 fields. Primary Key, Userid, Value1, Value2 Userid holds the User Id for the site. Value one is basically anything. It could be state, zip, phone etc. Value2 relates to Value1. I would like to make an sql query that would echo each UNIQUE Value2 and display a Count next to it which represents how many times it's duplicated. Example: CA(200) NY(300) TX(250) etc. Here's what I have so far. SELECT COUNT(DISTINCT value2) FROM mydb WHERE value1 = 'state' This returns 43 Which is probably correct because there are only 50 states in the U.S. Do I need a foreach loop? Please help. Thanks in Advance!

Sorry output is: <?xml version="1.0" encoding="utf-8"?><rss version="2.0"><channel><title>Headmine</title><description>Example of a RSS feed.</description><link>http://headmine.org</link><copyright>Copyright (C) 2009 headmine.org</copyright><item><enclosure length="1234567" type="image/jpeg" url="http://johnregalado.com/the-test/a-river-runs-though-it.jpg" /><url>http://headmine.org/the-test/a-river-runs-though-it.jpg</url></item><item><enclosure length="1234567" type="image/jpeg" url="http://johnregalado.com/the-test/mark-ruhi.jpg" /><url>http://headmine.org/the-test/mark-ruhi.jpg</url></item><item><enclosure length="1234567" type="image/jpeg" url="http://johnregalado.com/the-test/mark.jpg" /><url>http://headmine.org/the-test/mark.jpg</url></item><item><enclosure length="1234567" type="image/jpeg" url="http://johnregalado.com/the-test/namrok.jpg" /><url>http://headmine.org/the-test/namrok.jpg</url></item></channel></rss>

I've made a few edits.... Here is the link to the page: http://headmine.org/the-test/index.php I've only been able to get the icons to show with links. I'd like the images to display. here is the updated code <?php $channel = array("title" => "Headmine", "description" => "Example of a RSS feed.", "link" => "http://headmine.org", "copyright" => "Copyright (C) 2009 headmine.org"); foreach (glob("*.jpg") as $filename) { $items[] = array("img" => $filename); } $output = '<?xml version="1.0" encoding="ISO-8859-1"?>'; $output .= '<rss version="2.0">'; $output .= "<channel>"; $output .= "<title>" . $channel["title"] . "</title>"; $output .= "<description>" . $channel["description"] . "</description>"; $output .= "<link>" . $channel["link"] . "</link>"; $output .= "<copyright>" . $channel["copyright"] . "</copyright>"; foreach ($items as $item) { $output .= "<item>"; $output .= "<enclosure length=\"1234567\" type=\"image/jpeg\" url=\"http://johnregalado.com/the-test/". $item["img"] ."\" />"; $output .= "<url>http://headmine.org/the-test/". $item["img"] ."</url>"; $output .= "</item>"; } $output .= "</channel>"; $output .= "</rss>"; header("Content-Type: application/rss+xml; charset=utf-8"); echo $output; ?>

I don't understand? Like this foreach ($items as $item) { $output .= "<items>"; $output .= "<item>"; $output .= "<content><img src=\"http://LOCATION/" . $item["img_local"] . "\" /></content>"; $output .= "</item>"; $output .= "</items>"; }

When you view the file through a web browser it looks like a feed but no images show up. If I remove <img src= etc then the name of the files will appear. But as soon as I add img tags everything disappears.

What im trying to do is create a dynamic rss image feed based on images within a folder. Your array fix helped me but its . Still not working foreach (glob("*.jpg") as $filename) { $items[] = array("img_local" => $filename); } $output = '<?xml version="1.0" encoding="ISO-8859-1"?>'; $output .= '<rss version="2.0">'; $output .= "<channel>"; $output .= "<title>" . $channel["title"] . "</title>"; $output .= "<description>" . $channel["description"] . "</description>"; $output .= "<link>" . $channel["link"] . "</link>"; $output .= "<copyright>" . $channel["copyright"] . "</copyright>"; foreach ($items as $item) { $output .= "<item>"; $output .= "<content><img src=\"http://LOCATION/" . $item["img_local"] . "\" /></content>"; $output .= "</item>"; } $output .= "</channel>"; $output .= "</rss>"; header("Content-Type: application/rss+xml; charset=ISO-8859-1"); echo $output;

Is there an easier way to do this? I am trying to get create a dyamic array based on files within the folder. foreach (glob("*.jpg") as $filename) { $items = array("title" => $filename) } $output = '<?xml version="1.0" encoding="ISO-8859-1"?>'; $output .= '<rss version="2.0">'; $output .= "<channel>"; $output .= "<title>" . $channel["title"] . "</title>"; $output .= "<description>" . $channel["description"] . "</description>"; $output .= "<link>" . $channel["link"] . "</link>"; $output .= "<copyright>" . $channel["copyright"] . "</copyright>"; foreach ($items as $item) { $output .= "<item>"; $output .= "<title>" . $item["title"] . "</title>"; $output .= "</item>"; } $output .= "</channel>"; $output .= "</rss>"; header("Content-Type: application/rss+xml; charset=ISO-8859-1"); echo $output; ?> This doesn't even work correctly.

Never mind. I've asked about this before and just found my answer. Anyway to delete this?

Ok new problem... The form Needs to collect Company Name, and Contact Number(s). So Here is the form. <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <title>Insider Tips and Updates</title> <script> function addElement() { var ni = document.getElementById('myDiv'); var numi = document.getElementById('theValue'); var num = (document.getElementById('theValue').value -1)+ 2; numi.value = num; var newdiv = document.createElement('div'); var divIdName = 'my'+num+'Div'; newdiv.setAttribute('id',divIdName); newdiv.innerHTML = '<input name="aNumber[]" type="text" id="aNumber" size="12" /> <select name="aNumber[]"> <option value="Option">Option</option> <option value="Extension" selected="selected">Extension</option> </select> <input name="extOp[]" type="text" id="extOp" size="6" /><a href=\'#\' onclick=\'removeElement('+divIdName+')\'>Remove this number</a>'; ni.appendChild(newdiv); document.getElementByID('howMany').value = num; } function removeElement(divNum) { var d = document.getElementById('myDiv'); var olddiv = document.getElementById(divNum); d.removeChild(olddiv); } </script> </head> <body> <form id="form1" name="form1" method="post" action="process.php"> Company Name<br /> <label> <input name="company" type="text" id="company" /> </label><br /> Contact Numbers <br /> <div id="myDiv"></div> <label> <input type="submit" name="Submit" value="Submit"/> </label> </form> <input name="hidden" type="hidden" id="theValue" value="0" /> <p><a href="javascript:;" onclick="addElement();">Add contact number</a></p> <p> </p> <p> </p> </body> </html> If there is more than one contact number the user can click "ADD CONTACT NUMBER" and another dynamic field is added. There is an input for: Phone Number, Option for Extension/Option, and the Extension/Option Number. I added a drop down menu for the user to select weather they need an extension or option number. its named select[] So there are 3 different form variables being passed here. aNumber[], select[], and extOp[] So what I need the php script to do is create a string for each dynamic field that looks something like this $contact = aNumber[] . select[] . extOp[]; if there are more than one contact number then i'd like them to be added to $contact but seperated by comma's because they will be inserted into a database. So I tried $i = 0; foreach ($_POST["aNumber"] as $key) $i++; echo $i; $j = 0; while ($j<=$i) { $content = $_POST["aNumber"] . "," . $_POST["select"] . "," . $_POST["extOp"]; $j++; } echo $content; That didn't work. So I am stuck... Thanks in advance for any help

@mikr WOW! Thanks soo much for that! It does exactly what I need it to do! PERFECTION! and no I did not know about elements[] You've opened up a world of possibilities for me now! THANK YOU THANK YOU THANK YOU

This is driving me insane. Why does it remove ALL the commas? I only want it to remove the last character $string = substr($str, 0, -1); echo ($string); this echos a,b,c, $string = substr($str, 0, -2); echo ($string); this echos abc How do I only remove the last character?

@radi8 That didn't seem to do anything? @mikr What you did worked. But it duplicated 1 value this is how it printed out a,b,c,d,d d was not duplicated from the form so somewhere in the script it causes it to duplicate. Here is what I am running. foreach($_POST as $key => $value) if ($value == "Submit") { } else { if (strpos($key, "bp") !== false) { $theString = "$value,"; } #echo "$key: $value<br>"; echo $theString; } So it works! That's the first phase. But when inserting into the SQL database I need to remove that last "," how would I do that? I tried $newString = rtrim($theString, ","); echo ($newString); but that just removed ALL commas. Which I don't want.

Ok so I have a form that will dynamically add fields to insert more info. Here is the code for that. <form id="form1" name="form1" method="post" action="process.php"> Company Name<br /> <label> <input name="company" type="text" id="company" /> </label><br /> Bullet Points<br /> <div id="myDiv"></div> <label> <input type="submit" name="Submit" value="Submit"/> </label> </form> <input name="hidden" type="hidden" id="theValue" value="0" /> <p><a href="javascript:;" onclick="addElement();">Add bullet point</a></p> The Process.php file looks like this. foreach($_POST as $key => $value) if ($value == "Submit") { } else { if (strpos($key, "bp")) { $theString = "$value,"; } echo "$key: $value<br>"; echo $theString; } What I am trying to do is exclude submit as a value. The form key that is duplicated is bp followed by its number ex. bp1 = a bp2 = b bp3 = c etc I would like the values of only those keys to create a string. example $theString = "a,b,c"; I thought the above code would work. but it doesnt. What am I doing wrong? Also the idea behind this is to allow submission of a "Company Name" Once they enter the company name they can add bullet points to the form for the company. I figured the easiest way to do this is the above method since I never know how many bullet points will be needed. So the SQL statment would be something like "INSERT INTO company (uid, bulletPoints) VALUES('$uid', '$theString'" Am I going about this in the wrong way? Thanks in advanced for any help!

Here's a quick question.. Where can I read about how to condense my scripts? I got this entire thing working. Basically Page 1 displays all the images in the folder <?php $files = glob('images/*.jpg'); $count = 0; $counter = 0; foreach ($files as $file) { echo ($count % 4 == 0)? "<br/>" : ""; echo "<a href=\"img.php?file=$file&count=$counter\" /><img src=\"$file\" width=\"100\" style=\"border:1px solid #e96302;\" /></a>"; $count++; $counter++; } ?> Once you click on the image it will take you to a page to view the images. <div align="center"> <table width="425" border="0" align="center" cellpadding="0" cellspacing="0"> <tr> <?php if(empty($_GET['file'])) { header('Location:view.php'); } else { $ifile = $_GET['file']; $files = glob('images/*.jpg'); $count = $_GET['count']; $next = $count + 1; $prev = $count - 1; $taly = 0; foreach ($files as $file) { $taly++; } ?> <?php if($_GET['count'] == 0) { } else {?> <td width="22"><a href="<?php echo("img.php?file=$files[$prev]&count=$prev"); ?>"><img src='../../../img/prev.png' width='22' height='24' border="0" /></a></td> <?php }?> <td width="381"><div align="center"><a href="view.php"><img src='<?php echo $ifile; ?>' border="0" /></a></div></td> <td width="22"><?php if($_GET['count'] != $taly -1) { ?> <a href="<?php echo("img.php?file=$files[$next]&count=$next"); ?>"><img src='../../../img/next.png' width='22' height='24' border="0" /></a> <?php } } ?></td> </tr> </table> </div> Thanks to Crayon Violet the first code is nice and tight. The second code is sloppy i think. So are there any good articles on how to make scripts smaller?

Yes working great now! Thank you

Wow! Thanks crayon violet that didn't work at first but I made a few adjustments and it works perfect! Here is what I did $files = glob('images/*.jpg'); $count = 0; foreach ($files as $file) { echo ($count % 4 == 0)? "<br/>" : ""; $count++; echo "<a href=\"img.php?file=$file\" /><img src=\"$file\" width=\"100\" style=\"border:1px solid #e96302;\" /></a>"; } Basically just made $count = 0; and changed the path to the file because the way you had it printed out this way "/image/images/image.jpg" THANKS ALOT!

I found the problem Then problem was with this while($count <= $counter) { I switched it to while($count < $counter) { All errors are gone now. =) Thanks everyone for the help. Here is the final code for anyone reading along. <?php $count = 0; $counter = 0; $break = 0; $dir = opendir("images"); while (($file = readdir($dir)) !== false) { if(stristr($file, ".jpg")) { $name[$counter] = $file; $counter++; } } while($count < $counter) { echo "<a href=\"img.php?file=$name[$count]\" /><img src=\"images/$name[$count]\" width=\"100\" style=\"border:1px solid #e96302;\" /></a>"; $count++; $break++; if($break == 4) { echo '<br />'; $break = 0; } } closedir($dir); ?>

Thanks for the quick reply but that didn't do anything. It is still printing out the same error. Any other ideas?

Ok I included that in some changes I made to the script and that fixed the looping problem all images are displaying but now I am getting an error. Here is the code <?php $count = 0; $counter = 0; $break = 0; $dir = @ opendir("images"); while (($file = readdir($dir)) !== false) { if(stristr($file, ".jpg")) { $name[${"counter"}] = $file; $counter++; } } while($count <= $counter) { echo "<a href=\"img.php?file=$name[$count]\" /><img src=\"images/$name[$count]\" width=\"100\" style=\"border:1px solid #e96302;\" /></a>"; $count++; $break++; if($break == 4) { echo '<br />'; $break = 0; } } closedir($dir); ?> here is the error PHP Notice: Undefined offset: 15 in PATH\TO\FILE\view.php on line 26 PHP Notice: Undefined offset: 15 in PATH\TO\FILE\view.php on line 26 it prints out twice. Also I noticed that at the end something is coming up blank here is what the script is printing out <a href="img.php?file=Picture 108.jpg" /><img src="images/Picture 108.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 119.jpg" /><img src="images/Picture 119.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 120.jpg" /><img src="images/Picture 120.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 136.jpg" /><img src="images/Picture 136.jpg" width="100" style="border:1px solid #e96302;" /></a><br /><a href="img.php?file=Picture 138.jpg" /><img src="images/Picture 138.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 139.jpg" /><img src="images/Picture 139.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 140.jpg" /><img src="images/Picture 140.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 142.jpg" /><img src="images/Picture 142.jpg" width="100" style="border:1px solid #e96302;" /></a><br /><a href="img.php?file=Picture 143.jpg" /><img src="images/Picture 143.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 144.jpg" /><img src="images/Picture 144.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 145.jpg" /><img src="images/Picture 145.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 146.jpg" /><img src="images/Picture 146.jpg" width="100" style="border:1px solid #e96302;" /></a><br /><a href="img.php?file=Picture 147.jpg" /><img src="images/Picture 147.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 148.jpg" /><img src="images/Picture 148.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=Picture 149.jpg" /><img src="images/Picture 149.jpg" width="100" style="border:1px solid #e96302;" /></a><a href="img.php?file=" /><img src="images/" width="100" style="border:1px solid #e96302;" /></a><br /> Any help is appreciated.

I am trying to build my own gallery that reads the files from a folder. What I Need it to do is Choose only jpgs from the folder. Display 4 jpgs in a row the insert a line break to display the next for images Here is the code I have so far and it only displays one image over and over again and will not stop looping. <?php $count = 0; $counter = 0; $dir = @ opendir("images"); while (($file = readdir($dir)) !== false) { if(strpos($file, ".jpg")) { $name[${$counter}] = $file; $counter++; } } while($count <= $counter) { echo "<a href='img.php?file=". $name[${$count}]" /><img src='images/" . $name[${$count}] . "' width='100' style='border:1px solid #e96302;' /></a>"; $count++; if($count == 4) { echo '<br />'; $count = 1; } } closedir($dir); ?> im pretty sure the problem lies within the array. $name[${$counter}] = $file; Or has something to do with the $counter variable inside the name array. Please help! Thanks in advance