mysql nedod ārā datus

[nakedmika]

Sveiki!

nu jau kādu 3 stundu mokos ar vienu mazu knifu manā kodā. Gribu lai nospiežot pogu, man parādās dati no datubāzes, bet lai kā nespaidītos, man met ārā šo >>

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 ''navigacija' WHERE 'id' = 'undefined'' at line 1

 

un pats php kods ir šāds >>

<?php
$q=$_GET["q"];

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

mysql_select_db("portfolio", $con);

$sql="SELECT * FROM 'navigacija' WHERE 'id' = '".$q."'";

$result = mysql_query($sql) or die(mysql_error());

echo "<table border='1'>
<tr>
<th>id</th>
<th>name</th>
<th>content</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['content'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysql_close($con);
?>

[cucumber]

Nu tak lasi ko tev raksta

 

SELECT * FROM 'navigacija' WHERE 'id' = '".$q."'";

 

jabut

 

$q = mysql_escape_stirng($q);

SELECT * FROM `navigacija` WHERE `id` = '".$q."'"; /* savadakas pedinas */

[nakedmika]

nekas nemainās.. tas pats error. :(

bet vai tas ko manītu, ja piebilstu , ka poga darbojās ar ajax?

- pogas kods -

<div class="main">
<div id="list">
	<img src="sakums.png" alt="sākums" value="1" onclick="showUser(this.value)"/><br>
	<img src="darbi.png" alt="mani darbi" value="2" onclick="showUser(this.value)"/><br>
	<img src="kontakti.png" alt="kontakti" value="3" onclick="showUser(this.value)"/><br>
</div>
<div id="show">
		<!-- te nāk ajax sarazotie dati. -->
</div>
</div>

 

- ajax kods -

 

var xmlHttp;

function showUser(str)
{ 
xmlHttp=GetXmlHttpObject();
if (xmlHttp==null)
{
alert ("Browser does not support HTTP Request");
return;
}
var url="get.php";
url=url+"?q="+str;
url=url+"&sid="+Math.random();
xmlHttp.onreadystatechange=stateChanged;
xmlHttp.open("GET",url,true);
xmlHttp.send(null);
}

function stateChanged() 
{ 
if (xmlHttp.readyState==4 || xmlHttp.readyState=="complete")
{ 
document.getElementById("show").innerHTML=xmlHttp.responseText;
} 
}

function GetXmlHttpObject()
{
var xmlHttp=null;
try
{
// Firefox, Opera 8.0+, Safari
xmlHttp=new XMLHttpRequest();
}
catch (e)
{
//Internet Explorer
try
 {
 xmlHttp=new ActiveXObject("Msxml2.XMLHTTP");
 }
catch (e)
 {
 xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
 }
}
return xmlHttp;
}

 

- nu un protams jau augstāk minētais php kods -

 

<?php
$q=$_GET["q"];

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

mysql_select_db("portfolio", $con);

$sql="SELECT * FROM 'navigacija' WHERE 'id' = '".$q."'";

$result = mysql_query($sql) or die(mysql_error());

echo "<table border='1'>
<tr>
<th>id</th>
<th>name</th>
<th>content</th>
</tr>";

while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['content'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysql_close($con);
?>

[cucumber]

Es tak skaidri un gaishi pateicu kur ir errors SQL'a, ta ka tu uzrakstiji nestradas un vis

 

 

Otrs, izmanto php.lv/paste shitik lieliem koda blokiem

 

Treshkart, tev nemaz uz sql'u id salidzinashanas vertiba nenonak JS ar apskaties velreizs

Edited March 20, 2009 by cucumber

[Mr.Key]

SQL vai nu nelieto pēdiņas ap tabulu un lauku nosaukumiem vispār, vai arī lieto "backticks" (tās nav ne pēdiņas, ne apostrofs!)

 

SELECT * FROM table_name WHERE id = ...

SELECT * FROM `table_name` WHERE `id` = ...

/* Nepareizi */ SELECT * FROM 'table_name'

 

RTFM!!! RTFM!!! RTFM!!!

Edited March 20, 2009 by Mr.Key