грешка при заявка в база данни Syntax Error

[heji]

Заявката

<form action="insert.php" method="post">
Èìå: <input type="text" name="ime" />            <br>
Âåðñèÿ: <br>
<input type="checkbox" value="xp" NAME="ver[]">  WindowsXP        <br>
<input type="checkbox" value="vista" NAME="ver[]">      Windows Vista <br>
<input type="checkbox" value="7" NAME="ver[]">    Windows 7       <br>
×èïîâå: <input type="text" name="chips" /> <br>
Äîïúëíèòåëíà èíôîðìàöèÿ: <input type="text" name="descrip" />    <br>
<input type="submit" value="Äîáàâè"/>
</form>

пхп кода

<?php

$media_array = $_POST['ver'];
foreach ($media_array as $one_media) {
$source .= $one_media.", ";
}
$media = substr($source, 0, -2);


$con = mysql_connect("localhost","*****","******");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("pigeonz_iss", $con);

$sql="INSERT INTO 'drivers' ('name', 'ver', 'chip', 'desc')
VALUES ('$_POST[ime]', '$version', '$_POST[chips]', '$_POST[descrip]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Äðàéâåðà å óñïåøíî äîáàâåí!!!";

mysql_close($con)
?>

грешката

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''drivers' ('name', 'ver', 'chip', 'desc') VALUES ('Драйвер 1', '', '2234, 2342, ' at line 1

 

[Karcho]

<?php

$media_array = $_POST['ver'];
foreach ($media_array as $one_media) {
$source .= $one_media.", ";
}
$media = substr($source, 0, -2);


$con = mysql_connect("localhost","*****","******");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("pigeonz_iss", $con);

$sql="INSERT INTO `drivers` (`name`, `ver`, `chip`, `desc`)
VALUES ('$_POST[ime]', '$version', '$_POST[chips]', '$_POST[descrip]')";

if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Äðàéâåðà å óñïåøíî äîáàâåí!!!";

mysql_close($con)
?>

 

[heji]

Горепосоченият проблем го реших, но сега имам проблем със търсенето ето кода

<HTML>
<HEAD>
  <meta http-equiv="Content-Type" content="text/html; charset=windows-1251" />
 <TITLE>Òúðñåíå ïî ÷èï:</TITLE>
</HEAD>
<BODY>
<center>
Òúðñåíå ïî ÷èï:
<table border="1">
 <tr bgcolor="<?if(($count %2) == 0){echo'ffffff';}else{echo'd4e7ff';} $count++; ?>">
  <td cellspacing="0" cellpadding="0" align="center">No:</td>
  <td cellspacing="0" cellpadding="0" align="center">Èìå:</td>
  <td cellspacing="0" cellpadding="0" align="center">Windows:</td>
  <td cellspacing="0" cellpadding="0" align="center">×èïîâå:</td>
  <td cellspacing="0" cellpadding="0" align="center">Äîïúëíèòåëíî:</td>
  <td cellspacing="0" cellpadding="0" align="center"> </td>
 </tr>
<?
$con = mysql_connect("localhost","@@@@@@@@","@@@@@@");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
$chipa = $_POST["chip"];

mysql_select_db("pigeonz_iss", $con);

$result = mysql_query(" SELECT * FROM 'drivers' WHERE 'chips' LIKE '$chipa' ");

if($row = mysql_fetch_array($result))
  {
?>

 <tr bgcolor="<?if(($count %2) == 0){echo'ffffff';}else{echo'd4e7ff';} $count++; ?>">
  <td cellspacing="0" cellpadding="0" align="center"><? echo $row['id']; ?></td>
  <td cellspacing="0" cellpadding="0" align="center"><? echo $row['name']; ?></td>
  <td cellspacing="0" cellpadding="0" align="center"><? echo $row['ver']; ?></td>
  <td cellspacing="0" cellpadding="0" align="center"><? echo $row['chips']; ?></td>
  <td cellspacing="0" cellpadding="0" align="center"><? echo $row['desc']; ?></td>
  <td cellspacing="0" cellpadding="0" align="center"><a href="a">Èçòåãëè!</a></td>
 </tr>

<?}
mysql_close($con);
?>
</table>
</BODY>
</HTML>

изписва ми

Търсене по чип:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/pigeonz/public_html/iss/searchc.php on line 30
No: 	Име: 	Windows: 	Чипове: 	Допълнително:

 

[heji]

и това го оправих как да направя да показва съответствие дори и при част от фраза в барата данни?

 

[Karcho]

Направи заявката така:


$result = mysql_query(" SELECT * FROM `drivers` WHERE `chips` LIKE '%$chipa%' ");