вот мой код при его запуске выскакивает сообщение:
"The image “
http://172.17.0.43/avtobrest.com/find_car.php?marka=Audi&=m” cannot be displayed, because it contains errors."
$host = "localhost";
$user_db = "root";
$pass_db = "";
$dbase = "avtobrest";
mysql_connect ($host,$user_db,$pass_db);
mysql_select_db("$dbase");
$dtable = cars;
$marka = $HTTP_GET_VARS["marka"];
$model = $HTTP_GET_VARS["model"];
$year_from = $HTTP_GET_VARS["year_from"];
$year_to = $HTTP_GET_VARS["year_to"];
$cena_from = $HTTP_GET_VARS["cena_from"];
$cena_to = $HTTP_GET_VARS["cena_to"];
if($model != NULL)
{
$sql = "SELECT id_car FROM $dtable WHERE marka=\\"$marka\\" AND model=\\"$model\\" AND (year BETWEEN $year_from AND $year_to) AND (cena BETWEEN $cena_from AND $cena_to)";
$result = mysql_query($sql);
$rows = mysql_num_rows($result);
}
else
{
$sql = "SELECT id_car FROM $dtable WHERE marka=\\"$marka\\"";
$result = mysql_query($sql);
$rows = mysql_num_rows($result);
}
$i = 0;
if (!$rows)
{
echo "Данных нет";
}
else
{
echo"
";
while($row = mysql_fetch_array($result))
{
++$i;
echo"
$i .";
$id_car = $row["id_car"];
$marka = mysql_query("SELECT marka FROM $dtable WHERE id_car = \\"$id_car\\" ");
$marka = mysql_fetch_array($marka);
$marka = $marka[0];
$year = mysql_query("SELECT year FROM $dtable WHERE id_car = \\"$id_car\\" ");
$year = mysql_fetch_array($year);
$year = $year[0];
echo"$marka, $year г.в.
|
";
$foto1 = mysql_query("SELECT foto1 FROM $dtable WHERE id_car = \\"$id_car\\" ");
$foto1 = mysql_fetch_array($foto1);
$foto1 = $foto1[0];
header("Content-type: image/gif");
echo $foto1;
echo" |
";
$info = mysql_query("SELECT info FROM $dtable WHERE id_car = \\"$id_car\\" ");
$info = mysql_fetch_array($info);
$info = $info[0];
echo $info;
echo" |
|
";
}
echo"
";
}
?>
а если закоментировать строку header("Content-type: image/gif"); то выводится содержимое файла ввиде текста
как мне исправить ошибку???
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