php

I want to save the pictures (jpg files) in the database and i want to view that pictures using php. please help me. what is the code for that?

It's exactly the same as reading/writing anything else to/from a database. Make sure, though, that the row/column* you're writing it into allows binary input. In MySQL, this means it needs to be of the BLOB type with its BINARY attribute set.
Then, just store/retrieve it normally, and use a header("Content-Type: image/jpeg"); statement to set the MIME-type correctly before echoing the data to the browser.

* preferred visualisation varies depending on database package.

not to drag this off topic, but a quick question about the MIME-type command...

I saw that as an example for a page that was a php-generated image... and it's url was .../image.php, but it was just like a .jpg viewed as its own page.

Would you still need to include the MIME-type command if you were using the image in a page, such as the <img ...> command?

I might be missing something.

Quote:

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Originally Posted by djr33

I saw that as an example for a page that was a php-generated image... and it's url was .../image.php, but it was just like a .jpg viewed as its own page.

Would you still need to include the MIME-type command if you were using the image in a page, such as the <img ...> command?

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Yes. The user agent will (or at least should) depend upon the Content-Type HTTP header to determine how to handle a file. An accurate Content-Type header should be sent with every HTTP response that contains an entity body (returned content).

Mike

I solved that problem.
Another problem If I clike a button i want to view the data in another page.
(only using php)

Huh? Not too clear.

<form action='OTHERPAGE'>
<input type='submit'>
</form>

That would work. Put it inside an echo tag if you need it to be displayed by php.

...
echo "[insert above code]";
...

code is ok. But i want to pass a value to other page.
eg:- ______ search.
here i want to give a value (eg: name) for search & view the details of that data(eg:- age, firstname, lastname, etc...) another page using php.

Code:

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<form action='OTHERPAGE' method="post">
  <input type="text" name="search">
  <input type='submit' value="Search">
</form>

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On the PHP page, retrieve the value using

PHP Code:

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$_POST['search'] 


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I want to insert a picture inside the php code.
echo(" <tr> <td height=\"17\"><img src=\"a.jpg\" width=\"748\" height=\"20\"></td> </tr>");
is it correct?
But it is not working. why?

You don't insert a picture into PHP code; you cause PHP to output the HTML to display the picture.
That is correct. Is the code being called?

That coding is not working. Without php, only using html, it is working.

Are you certain the code is being called?