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I'm trying to print a list of links, this means I need to use a href in the 3rd line, but I don't know how the syntax should be in this situation. Also, the link is a PHP variable.
$sql = mysql_query("select nm_linha from linhas where cidades_origem like '%$cod_cidades%'");
while($exibe = mysql_fetch_assoc($sql)){
echo $exibe['nm_linha'] .'<br>';
}
$sql = mysql_query("select nm_linha from linhas where cidades_origem like '%$cod_cidades%'");
while($exibe = mysql_fetch_assoc($sql)){
echo "<a href='link.html'>".$exibe['nm_linha'] ."</a><br>";
}
$sql = mysql_query("select nm_linha from linhas where cidades_origem like '%$cod_cidades%'");
while($exibe = mysql_fetch_assoc($sql)){
echo '<a href='".$exibe['nm_linha']."'>'.$exibe['nm_linha'].'</a><br>';
}
this is going to list the links as well as , on clicking it is redirected to th respective link.
try this.
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I'm newbie so I hope u'll help me. I need to update my SQL with this PHP form. I googled a lot, but problem is i have name in variable:
($rowclanky["id"])
And simply just can't handle it... Tried a lot of things like adding slashes, etc. But I still couldn't make it. Can someone help me how to figure this out? Thanks a lot guys and sorry for my English.
<form method='POST'>;
$sql = "SELECT id,text FROM clanky";
$result = $con->query($sql);
if ($result->num_rows > 0){
while($row = $result->fetch_assoc()) {
echo "<input type='text' name=".$row["id"]." value='".$row["text"]."'><br>";
}
}
echo "<input type='submit' name='send' value='Upravit'><br> </form>;
}
You have not declared the $rowclacky variable. Try the following:
while ($row = $result->fetch_assoc()) {
echo "<input type='text' name=".$row["id"]." value='".$row["text"]."'><br>";
}
notice that you are setting the results from "fetch-assoc()" to the variable $row. So when you try to access the columns in you need to get them from the $row variable.
Hopefully this helps.
$rowclanky is not defined anywhere
$rowclanky["id"]
should be
$row["id"]
And
$rowclanky["text"]
should be
$row["text"]
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I have a problem with my code. I want to make a Seesion which display the current login user
Thank you !
$link = mysql_connect($DB_server, $DB_user, $DB_password,$DB_name);
mysql_select_db('linux', $link) or die(mysql_error());
$uid = 0;
if(isset($_GET["id"]))
{
$uid = $_GET["id"];
}
$profilname = mysql_query("SELECT U_Benutzername ,U_ID FROM u_user WHERE U_ID = '$uid'");
$emailadresse =mysql_query("SELECT U_login ,U_ID FROM u_user WHERE U_ID = '$uid'");
$_SESSION['profilname'] = $profilname;
$_SESSION['email'] =$emailadresse;
echo "Benutzername: $profilname";
echo "<br/>";
echo "Email: $profilname";
You have a typo...
$lid = $_GET["id"];
should probably be
$uid = $_GET["id"];
change your mysql query to:
if($uid!=0) {
$ref = mysql_query("SELECT * FROM u_user WHERE U_ID = '$uid'");
$row = mysql_fetch_array($ref);
$profilname = $row["U_Benutzername"];
$emailadresse= $row["U_login"];
$_SESSION['profilname'] = $profilname;
$_SESSION['email'] =$emailadresse;
}
to display it:
echo $_SESSION['profilname'];
BUT:
You wont get far with this, this is mysql_-code and its deprecated.
You have to use mysqli in newer PHP-Versions.
http://php.net/manual/de/book.mysqli.php
Change $lid = $_GET["id"]; to $uid = $_GET["id"];, and it also is a good idea to sanitize the data from $_GET,
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<?php if($_POST) {
$username = htmlspecialchars($_POST['username']);
$password = htmlspecialchars($_POST['password']);
$mysqli = new mysqli('localhost','root','','movie_posters');
$query = $mysqli->query("SELECT password FROM users WHERE username = '"$username"'");
} ?>
When I try this code on WAMP, I get error like; this http://i.stack.imgur.com/qcifR.jpg
What can I do ?
Do not use single and double quotes in your query.
This is the right way:
$query = $mysqli->query("SELECT password FROM users WHERE username = '$username'");
Otherwise, you will not print $username's value.
You have to put periods before and after you variable.
In your example:
$query = $mysqli->query("SELECT password FROM users WHERE username = '".$username."'");
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Hello guys i don't know what is call if or switch statement.
i have database table field which is call price, so i want to make if the price if the price field empty its show "please Call us for the price " and if the price field is not its i will show what ever i put it..
here is my code but i don't know why its not working..
any body can help please
thanks you
<?php
$result = mysql_query("SELECT * FROM articles where articlefriendlyURL='%s'",mysql_real_escape_string($aid));
while($row = mysql_fetch_assoc($result)) {
if ($row['price']; = = '') {
echo ("Please Call Us for the price");
else {
echo $row_getArticle['price'];
}
}
}
?>
You can't separate the equals signs or its no longer a conditional statement to check the equal to. you also had a syntax error with your if/else from the brackets. Try this:
<?php
$result = mysql_query("SELECT * FROM articles where articlefriendlyURL='%s'",mysql_real_escape_string($aid));
while($row = mysql_fetch_assoc($result)) {
if (empty($row['price'])) {
echo ("Please Call Us for the price");
} else {
echo $row['price'];
}
}
?>
I think the problem is on
if ($row['price']; = = '0')
I think the correct syntax of that statement is
if ($row['price'] == '0')
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<?php
//GET FACEBOOK LIKE DATA TO FACEBOOK SERVERS
function readFacebookLikes($url) {
$query = "select like_count from link_stat WHERE url ='" . $url ."'";
$s = file_get_contents("https://api.facebook.com/method/fql.query?query=".
urlencode($query)."&format=json");
preg_match("#(\"like_count\"):([0-9]*)#",$s,$ar);
if(isset($ar[2])) return $ar[2]; else return null;
}
mysql_connect('localhost','root','') or die(mysql_error());
mysql_select_db('odrinhasedm') or die(mysql_error());
$query=mysql_query("select nome, url from participacao") or die(mysql_error());
while($res=mysql_fetch_array($query))
{
$likes = "http://[ip]/evento-odrinhas-edm-open-air/".$res['url'];
$links = $res['url'];
$contador = readFacebookLikes($likes);
//echo'<tr><td>'.$res['nome'].'</td><td>'.$contador.'</td></tr>';
$query1 = "UPDATE participacao set likes = $contador WHERE url = $links ";
mysql_query($query1) or die (mysql_error());
}
?>
Why is this query1 not working? (PHPMyAdmin errror #1064)
I have a project that creates a page per user on a form and each page has its own Facebook like system, now I have a way to get that data and show it on a table, but I want to insert that data into my database to sort it. Is it easier to add the values to my db or sort the table directly in php?
I have a query that is working and its pulling the URLs to the PHP file, then I want to create another query to insert the data into the database with "update".
I think url is a string. So you must have quotes around the values:
UPDATE participacao set likes = $contador WHERE url = '$links'