1 Reply by Diwasas

(13 replies, posted in PHP)

nu man nereikia kad faila pati ikeltu i db o tik jo pavadinima, o failas butu kataloge kuri yra prie visu kitu php failu.

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2 Reply by Diwasas

(13 replies, posted in PHP)

<?php
$target = "foto"; 
$target = $target . basename( $_FILES['photo']['name']);
 $pav=$_POST['Pavadinimas']; 
 $aprasymas=$_POST['Aprasymas']; 
 $svoris=$_POST['Svoris']; 
 $kaina=$_POST['Kaina'];
 $amzius=$_POST['Amzius'];
 $foto=($_FILES['photo']['name']);
 $rusys=$_POST['Rusys']; 
 $lytis=$_POST['Lytis_id'];
 
$con = mysql_connect("localhost","xx","xxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("i8_j", $con);
$sql="INSERT INTO Gyvunai (Pavadinimas, Aprasymas, Svoris, Kaina, Amzius,Foto, Rusys_id, Lytis_id)
VALUES
('$pav','$aprasymas','$svoris','$kaina','$amzius','$foto', '{$rusys}', '{$lytis}')";
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
 { 
 echo "The file ". basename( $_FILES['uploadedfile']['name']). " ikeltas"; 
 } 
 else { 
 echo "Nepavyko ikelti failo."; 
 } 
mysql_close($con)
?>

prie to paties, neikelia foto. man reikia kad foto ikeltu i serveri mano o i db irasytu tik foto pavadinima, visad gaunu kad nepavyko ikelt

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3 Reply by Diwasas

(13 replies, posted in PHP)

Dekui uz pagalba :)

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4 Reply by Diwasas

(13 replies, posted in PHP)

:D nu tai savaime aisku, kad taip

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5 Reply by Diwasas

(13 replies, posted in PHP)

INSERT INTO Gyvunai (idGyvunai, Pavadinimas, Aprasymas, Svoris, Kaina, Amzius, Foto, Rusys_id, Lytis_id) VALUES ("","Zemaiciu
lenciuginis","istvermingas","30","20","6","null","","")

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6 Topic by Diwasas

(13 replies, posted in PHP)

Sveiki, darau duomenu forma apie gyvunus ir reikia padaryti radio buttonus, kurie i duomenu baze nusiustu value 1 arba 2.

<body>
<form action="men.php?page=insert.php" method="post">
<div align="center">
<br><br>
Veislė: <input type="text" align="middle" name="Pavadinimas" /></br>
Aprasymas: <input type="text" align="middle" name="Aprasymas" /></br>
Svoris: <input type="text" name="Svoris" /></br>
Kaina: <input type="text" name="Kaina" /></br>
Amzius: <input type="text" size="2" maxlength="2" name="Amzius" /></br>
Lytis: <input type="radio" name="Lytis_id" value="1">
       Patinas &nbsp; 
       <input type="radio" name="Lytis_id" value="2">
       Patele &nbsp; </br>
<br><br>
</div>
<input type="submit" />
</form>
</body>

insert.php

<?php
$con = mysql_connect("localhost","i8","Studi8");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("i8_juskevicius", $con);


$sql="INSERT INTO Gyvunai (Pavadinimas, Aprasymas, Svoris, Kaina, Amzius, Lytis_id)
VALUES
('$_POST[Pavadinimas]','$_POST[Aprasymas]','$_POST[Svoris]','$_POST[Kaina]','$_POST[Amzius]', '$_POST[Lytis_id]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Irasas pridetas!";

mysql_close($con)
?>

nelabai pagaunu esme kaip insert.php faile pertaisyt kad gautu id, nes man 0 irasi ir viskas. Aciu uz patarimus.

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