Проблемче в кода

spaidur

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Някои може ли да ми каже как може да се показва името место ID в линка на страницата ....
Код:
<?php
mysql_connect('localhost', 'root', 'pass');
mysql_select_db('db);

if(!$_GET['v']){
$sql = mysql_query('SELECT * FROM mp ORDER BY id DESC LIMIT 10');
while($row = mysql_fetch_assoc($sql))
{
echo '
<b><a href="meropriqtia.php?v='.$row['id'].'">'.$row['ime'].'</a></b><br />
';
?>
<?
}
}else{
$sql = mysql_query('SELECT * FROM mp WHERE id="'.(int)$_GET['v'].'"');
$row = mysql_fetch_array($sql);
 ?>
 <b><a href="meropriqtia.php?v=<?=$row['id']?>"><?=$row['ime']?></a></b><br />
 <?=$row['content']?>
 <?   
}
?>
 
Код:
<?php
mysql_connect('localhost', 'root', 'pass');
mysql_select_db('db);

if(!$_GET['v']){
$sql = mysql_query('SELECT * FROM mp ORDER BY id DESC LIMIT 10');
while($row = mysql_fetch_assoc($sql))
{
echo '
<b><a href="meropriqtia.php?v='.$row['name'].'">'.$row['ime'].'</a></b><br />
';
?>
<?
}
}else{
$sql = mysql_query('SELECT * FROM mp WHERE id="'.(int)$_GET['v'].'"');
$row = mysql_fetch_array($sql);
 ?>
 <b><a href="meropriqtia.php?v=<?=$row['name']?>"><?=$row['ime']?></a></b><br />
 <?=$row['content']?>
 <?   
}
?>
Там ли ? :roll:
 
idevbg каза:
Код:
<?php
mysql_connect('localhost', 'root', 'pass');
mysql_select_db('db);

if(!$_GET['v']){
$sql = mysql_query('SELECT * FROM mp ORDER BY id DESC LIMIT 10');
while($row = mysql_fetch_assoc($sql))
{
echo '
<b><a href="meropriqtia.php?v='.$row['name'].'">'.$row['ime'].'</a></b><br />
';
?>
<?
}
}else{
$sql = mysql_query('SELECT * FROM mp WHERE id="'.(int)$_GET['v'].'"');
$row = mysql_fetch_array($sql);
 ?>
 <b><a href="meropriqtia.php?v=<?=$row['name']?>"><?=$row['ime']?></a></b><br />
 <?=$row['content']?>
 <?   
}
?>
Там ли ? :roll:
да но така няма да стане имам в предвид ime кадето пише да излиза и линка и да си работи както трябва защото така ми се струва че ще даде грешка ..
 
Код:
<?php 
mysql_connect('localhost', 'root', 'pass'); 
mysql_select_db('db); 

if(!$_GET['v']){ 
$sql = mysql_query('SELECT * FROM mp ORDER BY id DESC LIMIT 10'); 
while($row = mysql_fetch_assoc($sql)) 
{ 
echo ' 
<b>
<a href="meropriqtia.php?v='.$row['id'].'&ime='.$row['ime'].'">'.$row['ime'].'</a></b><br /> 
'; 
?> 
<? 
} 
}else{ 
$sql = mysql_query('SELECT * FROM mp WHERE id="'.(int)$_GET['v'].'"'); 
$row = mysql_fetch_array($sql); 
 ?> 
 <b><a href="meropriqtia.php?v=<?=$row['name']?>"><?=$row['ime']?></a></b><br /> 
 <?=$row['content']?> 
 <?    
} 
?>
 
GodFather2 каза:
Код:
<?php 
mysql_connect('localhost', 'root', 'pass'); 
mysql_select_db('db); 

if(!$_GET['v']){ 
$sql = mysql_query('SELECT * FROM mp ORDER BY id DESC LIMIT 10'); 
while($row = mysql_fetch_assoc($sql)) 
{ 
echo ' 
<b>
<a href="meropriqtia.php?v='.$row['id'].'&ime='.$row['ime'].'">'.$row['ime'].'</a></b><br /> 
'; 
?> 
<? 
} 
}else{ 
$sql = mysql_query('SELECT * FROM mp WHERE id="'.(int)$_GET['v'].'"'); 
$row = mysql_fetch_array($sql); 
 ?> 
 <b><a href="meropriqtia.php?v=<?=$row['name']?>"><?=$row['ime']?></a></b><br /> 
 <?=$row['content']?> 
 <?    
} 
?>



Parse error: syntax error, unexpected T_STRING

Код:
<? 
} 
}else{ 
$sql = mysql_query('SELECT * FROM mp WHERE id="'.(int)$_GET['v'].'"'); 
$row = mysql_fetch_array($sql); 
 ?> 
 <b><a href="meropriqtia.php?v=<?=$row['name']?>"><?=$row['ime']?></a></b><br /> 
 <?=$row['content']?> 
 <?    
} 
?>
 

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