If PHP does not select an image, the selected image will be printed.

If you do not select an image in PHP, the front-end code of the selected image will be printed to view the image control.

while($row = mysql_fetch_array($result))  {echo'';  }

Background

For ($ I = 1; $ I <10; $ I ++) {$ j = $ I + 1; $ k = $ I + 5; if ($ _ FILES ["file '. $ I. '"] [" error "]> 0) {echo' no image selected '. $ j; $ {"iii ". $ I} = $ _ POST ['detail '. $ k];} else {if (file_exists ("images /". $ _ FILES ["file '. $ I. '"] [" name "]) {echo $ _ FILES [" file '. $ I. '"] [" name "]. 'image filename '. $ j. 'File name already exists; $ {"iii ". $ I} = '';} else {move_uploaded_file ($ _ FILES [" file '. $ I. '"] [" tmp_name "]," images /". $ _ FILES ["file '. $ I. '"] [" name "]); echo" Stored in :". "images /". $ _ FILES ["file '. $ I. '"] [" name "]; $ {" iii ". $ I} = 'Images /'. $ _ FILES ["file '. $ I. '"] [" name "] ;}}

Why do I not select an image and do not output it?

Reply to discussion (solution)

"File '. $ I .'"
This kind of thing cannot get the result you want, it should be
"File {$ I }"

You should change this first.

Isn't PHP variable separation all using '. $ I ..

I changed double quotation marks to single quotes.

Print_r ($ _ FILES) after submission; check what it is.