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Need help - Upload few images?


Hrvoje

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Hello...

 

I have some simple code but no ideas for upload another image??

PHP stores image name in database but there is no second picture uploaded in the images/ folder.

 

Pls help.

Thanks

 

if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2000000))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
    }
  else
    {
    echo "<strong>Slika:</strong> " . $_FILES["file"]["name"] . "<br />";
    echo "<strong>Format slike:</strong> " . $_FILES["file"]["type"] . "<br />";
    echo "<strong>Velicina:</strong> " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
    echo "<strong>Temp file:</strong> " . $_FILES["file"]["tmp_name"] . "<br />";



//  here is mistake???

      move_uploaded_file($_FILES["file"]["tmp_name"],
      "images/" . $_FILES["file"]["name"]);

//
      


$sql="INSERT INTO test (name, something, slika, slika2)
VALUES ('".$_POST['name']."', '".$_POST['something']."', '".$_FILES['file']['name']."', '".$_FILES['slika1']['name']."')";

if (mysql_query($sql))

{

    echo "Sucess";

} else {

    echo "Error" . mysql_error();

}
}
}
  else
  {
  echo "Invalid file";
  }
?>  

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https://forums.phpfreaks.com/topic/218781-need-help-upload-few-images/
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similar but little different try using this method:

 

$filename = $_FILES['upload']['name'];

if (move_uploaded_file($_FILES['upload']['tmp_name'], "../all_plants_img/$filename")) {

echo 'success image uploaded';

}

 

plus better to insert image info into db first then use

mysql_insert_id()

to get the int id and rename the image with the same number for easier reference

 

 

 

www.mymatesplace.com.au

Again my mistake sorry...

Does not work anymore:

 

Posting image name in dba but not upload images, only first image is uploaded .$_FILES['file']['name']

 

HTML input is 100% ok!

 

PHP:

 

<?php

include "database.php";




if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/png")
|| ($_FILES["file"]["type"] == "image/pjpeg"))
&& ($_FILES["file"]["size"] < 2000000))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
    }
  else
    {
    echo "<strong>Slika:</strong> " . $_FILES["file"]["name"] . "<br />";
    echo "<strong>Format slike:</strong> " . $_FILES["file"]["type"] . "<br />";
    echo "<strong>Velicina:</strong> " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
    echo "<strong>Temp file:</strong> " . $_FILES["file"]["tmp_name"] . "<br />";



    //   My old way...

    //   move_uploaded_file($_FILES["file"]["tmp_name"],
    //   "images/upload/" . $_FILES["file"]["name"]);
  


$filename = $_FILES['file']['name'];

if (move_uploaded_file($_FILES['file']['tmp_name'], "images/upload/$filename")) {

echo 'success image uploaded';

}

$sql="INSERT INTO artikli (ime_artikla, opis_artikla, grupacija, raspoloziv, slika, slika1, slika2, slika3, slika4, slika5)
VALUES ('".$_POST['ime_artikla']."', '".$_POST['opis_artikla']."', '".$_POST['grupacija']."', '".$_POST['raspoloziv']."', '".$_FILES['file']['name']."', '".$_FILES['slika1']['name']."', '".$_FILES['slika2']['name']."', '".$_FILES['slika3']['name']."', '".$_FILES['slika4']['name']."', '".$_FILES['slika5']['name']."')";

if (mysql_query($sql))

{

echo "Success";

} else {

echo "Error" . mysql_error();

}
}
}
  else
  {
  echo "Invalid file";
  }
?>

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