johnseito

Hello all, I was wondering if anyone can help me with putting this code in php. images is the directory, and John-Coltrane.jpg is the pic in the directory. So how can I do this if I am trying to get the img out of the directory onto the browser with php. would it be like this? Echo "<img src=images/$info['photo']/><br>"; images is the directory and $info['photo'] is the name of the img in the sql database, I did this and it doesn't work, any advise? thanks

Does it work for you? you talked as if you have it done

I did that, it didn't work.

I changed all the php.ini files to this and it still doesn't work Operating system ---- vista php version --------- PHP 5.2.3 apache version - -----http server 2.2.4 It just doesn't upload it, it pretty much gives me an error, because in my code if it doesn't do this if(move_uploaded_file($_FILES['uploaded']['tmp_name'], $target)) it will give me an error. When it upload successfully I will not see the error and I can see the file uploaded in the directory.

No, it doesn't work. any other ideas?

No, it doesn't work.

is there a way to upload excel docs or music docs that you know how to? I put excel docs like this but it doesn't work. thanks

But I do have a question if you know that would be great. I can only upload file size that are under 2mb, anything above that doesn't works. How can I do to change this, I changed the php ini to allow 300mb and it doesn't do any good. thanks

Ok I noticed something. I was missing these codes.. sometimes example codes are never completed. $uploaded_type = $_FILES['uploaded']['type'];

Here is what I have.

Yup, I changed it to how you have it and it still doesn't work.

I created a upload form to upload only a JPEG file type but it doesn't seem to work even if the file I upload is JPEG. Quote: if (($uploaded_type=="image/jpeg")) { anyone know what is wrong here? thanks

Thanks for the site, it just explains everything, doesn't really tells me what to do or how to fixed this problem. it even show me this code http://www2.radinks.com/upload/examples/upload.php.txt I load this up and I don't see anything.

Hello all, I created a form and a upload php to upload files in a directory but it seems that I can only upload files to a certain size limit, when files are 20mb it doesn't seem to do it. I went to php and set the upload size to 300mb and still it doesn't do it, but I could do this in my hosting company's directory and using their php but it just takes longer. why is that and how can I do this locally without it failing? thanks

Hello all, I created a form and a upload php to upload files in a directory but it seems that I can only upload files to a certain size limit, when files are 20mb it doesn't seem to do it. I went to php and set the upload size to 300mb and still it doesn't do it, but I could do this in my hosting company's directory and using their php but it just takes longer. why is that and how can I do this locally without it failing? thanks

Ok, thanks for your response. I created a table in MySQL called pix like this: Create table pix ( Pid int primary key not null auto_increment, Title text, Imgdata longblob); I know how to store the photo correctly into the pix table, but I couldn’t get all the uploaded photo out. I could only get one, the lastest one uploaded. And I couldn’t upgrade the code to do just that because of this: if ($_REQUEST[gim] == 1) { header("Content-type: image/jpeg"); print $bytes; exit (); } <img src=?gim=1 width=1000 height=1000><br> The gim==1 is what determines the photo to appear. I don’t know why is it gim ==1. So could you tell me why is src gim =1 or how I can change this code to show all photo uploaded instead of just 1. Thanks -

Hello all, People say that this code is suppose to work: while($row = mysql_fetch_array($query)) { echo "<tr>"; echo "<td><img src =\"" . $row['user_image']."\"></td>"; echo "</tr>"; it should bring out all your images/photo. I tried this code and no photo appear, just the icon, and when you upload one photo it seems that 3 icon will pop out as if there is 3 photo for every upload. but when I do it like this, Code: $row = mysql_fetch_assoc($query); $bytes = $row[imgdata]; $row = mysql_fetch_assoc($query); $bytes = $row[imgdata]; <center><img src=?gim=1 width=1000 height=1000><br> <!--144--> it works but it only brings out one photo. I want it to bring out all photo, imgdata is the field in mysql that stores the pictures. any help would be great. thankst

Ok I got that, now a couple of things if you could offer some advice on. 1- I have the codes to upload pix and show them in the browser, but in only does it for one pix at a time and the last one uploaded. I would like to show all the pixs in the database with the last one uploaded showed last, I tried for the past two days to do this but haven't found any solution. if you have any let me know. 2 - do you know if or the codes to upload music and excel files to the database and show them on the browser? I was thinking that maybe I should do it as I did it for pix and change it to xls for excel docs, etc. thanks

Hi I just noticed why some picture can't be uploaded, because some of those pictures files size are too large, about 600kb or 1.2 mb. I currently can only upload about 17 kb, how can I increase this capacity so that I can upload all my pictures? thanks

I want to show the files in the browser, the image of it. I created one and it works but it doesn't work for all pictures, it doesn't show all pictures only some. Do you know why that is? I also would like to upload music and ms doc (mostly excel) files, so how would I select these and view them on the browser?

cooldude832 thanks for the site Ok if it can store up to 4gb, how can i make it store more if in the end I want to? and I read the site provided by rhodesa, it said blob only stores . rhodesa thanks for the site, I will read more about it.

Thanks.. so far I have no clue what this field is. Can you show me a link that would give me more information on this?

Hello everyone, I have a question with upload files. I create a html form and php codes to upload files. I also created a directory called uploads where my php upload files resides. when I upload it, all the files goes to the uploads directory. so my question is when I upload can it be save into mysql and then I can bring it out and show it in the browser instead of how i have it in the directory. How would I do this? I really want to show what I upload in the browser/server. thanks -

I wrote this code $conn =mysql_connect("localhost","TEST1","TEST"); mysql_select_db("test",$conn); if (!(isset($pagenum))) /*isset() functions for testing the status of a variable. This function takes a variable name as an argument and returns true if it exists and false otherwise. you can also pass in a comma separated list of variables, and isset() will return true if all the varaibles are set. */ { $pagenum = 1; } $data ="SELECT name, message from info"; $rows = mysql_num_rows($data, $conn); print"<font size=10000>$rows</font>"; //$rows=mysql_query($data, $conn); $page_rows = 4; $last = ceil($rows/$page_rows); if ($pagenum < 1) { $pagenum = 1; } elseif ($pagenum > $last) { $pagenum = $last; } $max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows; $data_p = mysql_query($data, $conn) or die(mysql_error()); while ($info = mysql_fetch_array($data_p)){ echo "{$info['name']} wrote:<br>"; echo "{$info["message"]}"; } echo "<p>"; echo " --Page $pagenum of $last-- <p>"; // First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page. if ($pagenum == 1) { } else { echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=1'> <<-First</a> "; echo " "; $previous = $pagenum-1; echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$previous'> <-Previous</a> "; } //just a spacer echo " ---- "; //This does the same as above, only checking if we are on the last page, and then generating the Next and Last links if ($pagenum == $last) { } else { $next = $pagenum+1; echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$next'>Next -></a> "; echo " "; echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$last'>Last ->></a> "; } ?> <HTML> <HEAD> <TITLE>WEB BLOG</TITLE> </HEAD> <BODY> <H1>Add an Entry</H1> <form method="POST" action="pagination2.php"> <p> <input type="text" name="name" value= "rrrr" size=55><br> <b><small>NAME:</small></b><br></p> <b>MESSAGE:</b><br> <textarea name="message" cols="60" rows="6"></textarea> <br> <input type="submit" name="submit" value="Submit"> </form> </BODY> </HTML> it doesn't seem to work as plan. there are the -Page 0 of 0-- <<-First <-Previous ---- link to click on but that is it, I click first/previous and it does nothing, also it shows all the data output. Anyone have any ideas what is wrong here. thanks

thanks i'll take a look at it.