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Closed 9 years ago.
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Can someone help me to find what is wrong in the $secret line ?
$secret should give :
{"name":"JustAname","extra":"1","password":"ASD123","report":"http:\/\/website.com\/dev\/gamereport\/0001.php"}
here's the PHP code:
<?php
date_default_timezone_set('America/Montreal');
$name = 'JustAname';
$extra = '1';
$password = 'ASD123';
$reception = 'http:\/\/website.com\/dev\/gamereport.php';
// Code de génération de la base64
$secret = '{"name":"'.$name'","extra":"'.$extra'","password":"'.$password'","report":"'.$reception'"}';
$encodedSecret = base64_encode($secret);
$tournementLink = 'pvpnet://lol/customgame/joinorcreate/map1/pick6/team5/specALL/'.$encodedSecret;
echo $tournementLink;
?>
I got: Parse error: syntax error, unexpected T_CONSTANT_ENCAPSED_STRING in [...] on line 20
You're incorrectly concatenating strings, as #hobbs suggests. You're also using the undefined variable $Tournament, which I think should be $name. Try this:
$secret = '{"name":"' . $name . '","extra":"' . $extra . '","password":"' . $password . '","report":"' . $reception . '"}';
On a side note, a slightly nicer way to create JSON in PHP is to use an array and json_encode():
$secret = json_encode(array(
'name' => $name,
'extra' => $extra,
'password' => $password,
'report' => $reception));
Related
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Closed 3 years ago.
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I have this error when I run the code:
Error:
Invalid parameter number: number of bound variables does not match number of tokens - Line: 106
Code:
$data_cadastro = date("Y-m-d G:i:s");
$query = "INSERT INTO FRETES (VENDA_CLIENTE_ID_CLIENTE, VENDA_ID_VENDA, DT_COLETA, DT_ENTREGA, LINK, TRANSPORTADORA, POSICAO, VALIDA, DT_CADASTRO)
VALUES (:id_cliente, :id_venda, ':dt_coleta', ':dt_entrega', ':link', ':transportadora', ':posicao', :validacao, ':dt_cadastro')";
$banco = $this->pdo->prepare($query);
try {
$banco->execute(
array(
':id_cliente' => $this->id_cliente,
':id_venda' => $this->id_venda,
':dt_coleta' => $dados['DTcoleta'],
':dt_entrega' => $dados['DTentrega'],
':link' => $dados['linkFrete'],
':transportadora' => $dados['transportadora'],
':posicao' => $dados['posicaoFrete'],
':validacao' => $dados['validacao'],
':dt_cadastro' => $data_cadastro
)
);
} catch (PDOException $exception) {
die("Execução da Query com erro (inserir novo frete): " . $exception->getMessage() . ' - Linha: ' . $exception->getLine());
}
Where i wrong?
Remove the quotes around the placeholders
$query = "INSERT INTO FRETES
(VENDA_CLIENTE_ID_CLIENTE, VENDA_ID_VENDA, DT_COLETA,
DT_ENTREGA, LINK, TRANSPORTADORA, POSICAO, VALIDA, DT_CADASTRO)
VALUES (:id_cliente, :id_venda, :dt_coleta, :dt_entrega, :link,
:transportadora, :posicao, :validacao, :dt_cadastro)";
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Closed 7 years ago.
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I have a PDO PHP file making use of just one $_POST value stored on the $data array, and if an statement is true, a second value is added to that array to make a new query with two values:
<?php
session_start();
include("../conexionbbdd.php");
if($_SESSION['estado'] == 'activo' && $_SESSION['rol'] == '1'){
$data = array(
'us_id' => $_POST['us_id'],
);
$selectUsers= "SELECT * FROM ws_users WHERE us_id= :us_id";
$statementSelectUsers = $pdo->prepare($selectUsers);
$statementSelectUsers->execute($data);
$result = $statementSelectUsers->fetch(PDO::FETCH_ASSOC);
$us_fk_ui_id = $result['us_fk_ui_id'];
if($us_fk_ui_id==='1'){
$data['us_credits']=$_POST['us_credits'];
$updateUser = mysqli_query($con,"UPDATE ws_users SET us_credits = :us_credits, us_access = '1' WHERE us_id = :us_id");
$statementUpdateUser = $pdo->prepare($updateUser);
$statementUpdateUser->execute($data);
}
Everything goes fine untill the $statementUpdateUser->execute($data); line (34), where I get the usual error
PDOException: SQLSTATE[42000]: Syntax error or access violation: 1065
Query was empty in C:\wamp\www**********\actions\ad_updateUserInfo.php on
line 34
As far as I've seen, this should be due to the unexistance of one of the placeholders on the array, but if I print the array values after the $data['us_credits']=$_POST['us_credits']; it seems to be correct, having the 2 expected values needed for my query:
Array (
[0] => 2
[1] => 1.5 )
How could I check where the mistake is? There's no possibility of echoing the query as it is an object unable to transform on string.
$updateUser = mysqli_query($con,"UPDATE ws_users SET us_credits = :us_credits, us_access = '1' WHERE us_id = :us_id");
^^^ WTF??
You have to pay more attention to the code you write. Stack Overflow is NOT the service for finding typos for you.
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Closed 8 years ago.
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There is an error in my code but the code works perfectly. I mean all the values are inserted in the database but there is an error like this on the screen:
Severity: Notice
Message: Array to string conversion
Filename: models/some_model.php
Line Number: 106
This is my code:
View:
<?php foreach($app as $row){
echo "<tr><td><input type=checkbox name=appname[] value='".$row->app_name."'/>".$row->id."</td><td>".$row->app_name."</td><tr>".
?>
Controller:
public function hide(){
$this->load->model('some_model');
$visi = $this->input->post('appname');
$success = $this->some_model->hideApp($visi);
foreach($visi as $key=>$value)
{
$success = $this->some_model->hideApp($visi[$key]);
}
if($success == TRUE)
$this->hideApp_page(TRUE);
else $this->hideApp_page(FALSE);
}
Model:
public function hideApp($visi){
$visi = $this->db->escape_str($visi);
$queryStr = "UPDATE appwarehouse.application_table SET visibility='hidden' where app_name='$visi';"; /* this is line 106*/
$query = $this->db->query($queryStr);
return $query;
}
$visi is array like [1,2,3,4]
when you put $visi in hideApp()
it will be display "array to string error"
so maybe you can remove this line $success = $this->some_model->hideApp($visi);
you have already do some_model->hideApp($visi[$key]) in the foreach loop
so i don't know why you write this$success = $this->some_model->hideApp($visi);
if you still want to run $success = $this->some_model->hideApp($visi);
you have to put $visi into a string
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Closed 8 years ago.
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I'm new here and i'm an absolute beginner.
Please help with that error:
Parse error: syntax error, unexpected T_BOOLEAN_OR in index.php on line 174
First line is line 174
( $sql = || print mysql_error( ) );
if (0 < mysql_num_rows( $sql )) {
while ($row = mysql_fetch_assoc( $sql )) {
#extract( $row );
$numview = ($viewer == 1 ? 'view' : 'views');
$numvideo = ($counter == 1 ? 'video' : 'videos');
$chanlink = 'index.php?channels=browse&channel_id=' . $uid;
if (!empty( $img )) {
$show_images = '<img src="thumbs/channel/' . $img . '" width="' . $chan_img_width . '" height="' . $chan_img_height . '" border"0">';
}
else {
$show_images = '';
}
The error is pretty clear...
unexpected T_BOOLEAN_OR in index.php on line 174
This is line 174:
( $sql = || print mysql_error( ) );
This is a boolean OR:
||
You can't assign a boolean OR operator to a variable, so this doesn't make sense (let alone the rest of the line):
$sql = ||
The PHP parser doesn't expect things that don't make sense, so it throws an error.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 9 years ago.
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I need to format Date time to specific date format using php.
Now date format
2014-02-10 09:32:24
I need format
2014/02/10
error is :Parse error: syntax error, unexpected '$ModifiedDate' (T_VARIABLE), expecting ',' or ';' in ................. line
Here i tried like this
<?php
include(config.php);
$sql = "SELECT Id,Title,description,Image,Category,ModifiedDate from News WHERE ISActive=true";
$query = mysql_query($sql);
if(mysql_num_rows($query)>0){
while($test = mysql_fetch_array($query))
{
$Mod = $test['ModifiedDate'];
$ModifiedDate = date_format($Mod, 'yyyy/MM/dd');
echo"<td><font color='black'>"$ModifiedDate"</font></td>";
}
}
else
{
echo "No Records";
}
?>
You forgot to concatenate the variable. Do like this
echo"<td><font color='black'>".$ModifiedDate."</font></td>";
//^----- ^----- Add those dots !
Use below format,
date_format($Mod, 'YY/MM/DD');
try this code:
$ModifiedDate = date_format($Mod, 'Y/m/d');