PHP,MySql, resim upload ve bilgileri veritabanına gönderme?

Zapatas · 13 Ara 2009

Merhabalar arkadaşlar,

Resim upload eden ve bu resime ait bilgileri database e yerleştiren bir script yazdım ama sorunla karşılaşıyorum.

Sorunum, resim bilgilerinden resimin adı veritabanına doğru giriyor ama age ve gender 0(varsayılan değer) olarak giriyor, bense formdan aldığım değerleri girmek istiyorum.

Bunun dışında kodu daha da iyileştirmek için önerileriniz varsa onları da yazarsanız sevinirim.

Teşekkürler..

imageupload.htm

HTML:

<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<form action="imageupload.php" method="post" enctype="multipart/form-data" name="imageupload">
<table width="384" height="195" border="0" align="center">
  <tr>
    <td width="121" height="50">select picture</td>
    <td width="12">:</td>
    <td width="237"><input name="image" type="file" /></td>
  </tr>
  <tr>
    <td height="46">age</td>
    <td>:</td>
    <td><select name="age">
      <option value="14">14</option>
      <option value="15">15</option>
      <option value="16">16</option>
      <option value="17">17</option>
      <option value="18">18</option>
      <option value="19">19</option>
      <option value="20">20</option>
      <option value="21">21</option>
      <option value="22">22</option>
      <option value="23">23</option>
      <option value="24">24</option>
      <option value="25">25</option>
      <option value="26">26</option>
      <option value="27">27</option>
      <option value="28">28</option>
      <option value="29">29</option>
      <option value="30">30</option>
      <option value="31">31</option>
      <option value="32">32</option>
      <option value="33">33</option>
      <option value="34">34</option>
      <option value="35">35</option>
      <option value="36">36</option>
      <option value="37">37</option>
      <option value="38">38</option>
      <option value="39">39</option>
      <option value="40">40</option>
      <option value="41">41</option>
      <option value="42">42</option>
      <option value="43">43</option>
      <option value="44">44</option>
      <option value="45">45</option>
      <option value="46">46</option>
      <option value="47">47</option>
      <option value="48">48</option>
      <option value="49">49</option>
      <option value="50">50</option>
    </select>    </td>
  </tr>
  <tr>
    <td height="44">gender</td>
    <td>:</td>
    <td><select name="gender">
      <option value="female">Female</option>
      <option value="male">Male</option>
    </select></td>
  </tr>
  <tr>
    <td colspan="3"><div align="center">

        <label> <br />
        <input type="submit" name="submit" id="bas" value="Submit" />
          <br />
          </label>
      
    </div></td>
  </tr>
</table>

</form>





</body>
</html>

imageupload.php

PHP:

<?php 
include("baglanti.php"); 
if($_POST['image'] = "" || $_POST['gender'] = "" || $_POST['age'] = "") {
echo "Fill all blanks";
}else{
$image=$_POST['image'];
$gender=$_POST['gender'];
$age=$_POST['age'];
$imagename=$_FILES["image"]["name"];


//Dosya yükleme

if ((($_FILES["image"]["type"] == "image/gif")
|| ($_FILES["image"]["type"] == "image/jpeg")
|| ($_FILES["image"]["type"] == "image/pjpeg"))
&& ($_FILES["image"]["size"] < 200000))
  {
  if ($_FILES["image"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["image"]["error"] . "<br />";
    }
  else
    {
    echo "Upload: " . $_FILES["image"]["name"] . "<br />";
    echo "Type: " . $_FILES["image"]["type"] . "<br />";
    echo "Size: " . ($_FILES["image"]["size"] / 1024) . " Kb<br />";
    echo "Temp file: " . $_FILES["image"]["tmp_name"] . "<br />";

    if (file_exists("upload/" . $_FILES["image"]["name"]))
      {
      echo $_FILES["image"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["image"]["tmp_name"],
      "upload/" . $_FILES["image"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["image"]["name"];
      }
    }
  }
else
  {
  echo "Invalid file";
  }


// Veri girişi
$sql="INSERT INTO resim_upload (picture_name, gender, age)
VALUES ('$imagename','$gender','$age')";

if (!mysql_query($sql,$bag))
  {
  die('Error: ' . mysql_error());
  }
echo "1 record added";

}
?>

veritabanı yapısı ise

Kod:

rid                    mediumint(9)
picture_name     mediumtext
gender              bit(1)
age                  tinyint(4)

PHP,MySql, resim upload ve bilgileri veritabanına gönderme?

Zapatas

New member