random image help

@aytakinAug 23.2004

I have the following Random image PHP script, but it is not working.
“images/falehafez” is the directory of my images.
Please help me, and I’m not really a pro on PHP so please don’t use professional language. THANK YOU.

Starts:
<?
// #### RANDOM PIC #########################################

$thumbstring = ”; $file_dir=”images/falehafez”; // DIRECTORY WITH THE PICS

$f_type=”.jpg”; // FILE EXTENSION YOU WISH TO DISPLAY

$dir=opendir(images/falehafez); while ($file=readdir($dir)) { if ($file != “.” && $file != “..”) { $extension=substr($file,-3); // THIS DIGIT MUST MATCH THE NUMBER OF CHARACTERS SPECIFIED IN THE FILE EXTENSION ABOVE if($extension == $f_type) { $thumbstring .= “$file|”; } } } srand((double)microtime()*1000000); $arry_txt = explode(“|” , $thumbstring); echo “<img src=””images/falehafez”/”.$arry_txt[rand(0, sizeof($arry_txt) -1)].””>

Ends

to post a comment

PHP

8 Comments(s) _↴

@Paul_JrAug 23.2004 — #Try this.

[code=php]
 <?php
 // #### RANDOM PIC #########################################
 
 $thumbstring = '';
 $file_dir = 'images/falehafez'; // DIRECTORY WITH THE PICS
 $f_type = '.jpg'; // FILE EXTENSION YOU WISH TO DISPLAY
 if(($dir = @opendir($file_dir) !== false)) {
 while ($file = @readdir($dir)) {
 if ($file != '.' && $file != '..') {
 $extension = substr($file, -4); // THIS DIGIT MUST MATCH THE NUMBER OF CHARACTERS SPECIFIED IN THE FILE EXTENSION ABOVE
 if($extension == $f_type) {
 $thumbstring .= $file . '|';
 }
 }
 }
 }
 srand((double)microtime()*1000000);
 $arry_txt = explode('|' , $thumbstring);
 echo '<img src="' . $file_dir . '/' .  $arry_txt[rand(0, count($arry_txt) -1)] . '" />';
 ?>
 [/code]

Note: I'm not on my computer, so I can't test this out.

@aytakinauthorAug 23.2004 — #Thanks for your help, but it didn't work.

@Paul_JrAug 24.2004 — #Did you get any error messages or anything?

[code=php]
 <?php
 $thumb_string = array();
 $img_dir = '/path/to/images';
 if(is_dir($img_dir)) {
 if(@opendir($img_dir)) {
 while(($file = @readdir($img_dir)) !== false) {
 if($file != '.' && $file != '..' && substr($file, -4) == '.jpg')
 $thumb_string[] = $file;
 }
 }
 }
 }
 $rand_img = $thumb_string[rand(0, count($thumb_string) - 1)];
 echo '<img src="' . $img_dir . '/' . $rand_img . "' />';
 ?> 
 [/code]

Again, not on my computer, so I can't test it.

@aytakinauthorAug 24.2004 — #It again didn't work, but thanks.

@pyroAug 24.2004 — #You could try [url=http://www.alistapart.com/articles/betterrotator/]this one[/url].

@aytakinauthorAug 27.2004 — #Thanks,

but i'm looking for something that takes the pic from a directory, i have too many pics to add a link for each one.

@rrfuertesNov 13.2011 — #Try to use this. This code selects three unique random images from database (which is only the name of the picture) then it will output it on an html by comparing the name of pictures to the pictures in your directory.

[code=php]
 <?php
 $db = mysql_connect("localhost", "root", "");
 mysql_select_db("dbase1",$db);
 $range_result = mysql_query( " SELECT MAX(id) AS max_id , MIN(id) AS min_id FROM image1 ");
 $range_row = mysql_fetch_object( $range_result );
 $random1 = mt_rand( $range_row->min_id , $range_row->max_id );
 $result = mysql_query( " SELECT * FROM image1 WHERE id >= $random1-2");
 $img = mysql_fetch_array($result);
 $img1 = $img + 1;
 $img2 = $img + 2; 
 ?>[/code]

[code=html]
 <html>
 <body>
 <div>
 <li id="1">
 <?php
 echo '<img src='. "index3_files/{$img['name']}.jpg". ' alt="" height="438" width="700">';
 ?>
 <p><span><?php echo $img['caption']; ?></span></p>
 </li>
 <li id="2">
 <?php
 echo '<img src='. "index3_files/{$img1['name']}.jpg". ' alt="" height="438" width="700">';
 ?>
 <p><span><?php echo $img1['caption']; ?></span></p>
 </li>
 <li id="3">
 <?php
 echo '<img src='. "index3_files/{$img2['name']}.jpg". ' alt="" height="438" width="700">';
 ?>
 <p><span><?php echo $img2['caption']; ?></span></p>
 </li>
 </ul>
 </div>
 </body>
 </html>
 [/code]

@TecBratNov 15.2011 — #I have the following Random image PHP script, but it is not working.

"images/falehafez" is the directory of my images.

Please help me, and I'm not really a pro on PHP so please don't use professional language. THANK YOU.

Starts:

<?

// #### RANDOM PIC #########################################

$thumbstring = '';
  $file_dir="images/falehafez";	// DIRECTORY WITH THE PICS
 
  $f_type=".jpg";	// FILE EXTENSION YOU WISH TO DISPLAY
 
  $dir=opendir(images/falehafez);
  while ($file=readdir($dir))
  {
   if ($file != "." && $file != "..")
   {
   $extension=substr($file,-3);	// THIS DIGIT MUST MATCH THE NUMBER OF CHARACTERS SPECIFIED IN THE FILE EXTENSION ABOVE
   if($extension == $f_type)
   {
   $thumbstring .= "$file|";
   }
   }
  }
  srand((double)microtime()*1000000);
  $arry_txt = explode("|" , $thumbstring);
  echo "<img src=""images/falehafez"/".$arry_txt[rand(0, sizeof($arry_txt) -1)]."">

Ends[/QUOTE]

Looks like the error is in the last statement, the "echo". And I think that's what Paul Jr. was working on.

I think it should be like this:[code=php]echo '<img src="images/falehafez/'.$arry_txt[rand(0, sizeof($arry_txt) -1)].'">';[/code]

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