i have this code here. it gives me an error of Syntax error, unexpected T_STRING, expecting ',' or ';'. i want to be able to display a success message on the same page rather than opening another page. please help.
} else {
$query = "SELECT title, author, post, id FROM news_posts WHERE id=$id";
$result = mysql_query($query);
$num = mysql_num_rows($result);
$row = mysql_fetch_array ($result, MYSQL_NUM);
$title = $row['0'];
$name = $row['1'];
$message = $row['2'];
if ($num == 1) {
echo '
<form action="<?php $_SERVER['PHP_SELF']; ?>" method="post">
<p><b>Post Title :</b><br />
<input type="text" class="form-control" name="title" value="'.$title.'" /></p>
<p><input type="hidden" name="name" size="15" maxlength="255" value="'.$name.'" /></p>
<p><b>Post Message :</b><br /><textarea rows="7" class="form-control" name="message">'.$message.'</textarea></p>
<p><input type="button" value="Back" onclick="history.back()"> <input type="submit" name="submit" value="Submit" /></p>
<input type="hidden" name="submitted" value="TRUE" /></p>
<input type="hidden" name="id" value="'.$id.'" /></form>';
} else {
echo 'News post could not be edited, please try again.';
}
}
?>
It's because the single quotes around PHP_SELF are interrupting the single quotes in your echo. Try this instead:
<form action="' . $_SERVER['PHP_SELF'] . '" method="post">
Related
I'm trying to use both $_GET and $_POST in an sql query. The following is my code:
<?php
$assignment = mysql_real_escape_string($_GET['name']);
echo "$assignment <br>";
if (isset($_POST['add'])) {
$user = $_POST['username'];
$text = $_POST['comment'];
$query = "INSERT INTO comments (user, text, assignment) VALUES ('$user', '$text', '$assignment')";
mysql_query($query) or die('Error, comment failed to post');
}
?>
<h1>Add Comment</h1>
<form action="log_entry.php" method="post">
Name:<br/>
<input type="text" name="username" value="" />
<br /><br />
Comment:<br />
<textarea style="height:200px;" type="text" name="comment" value="" ></textarea>
<br /><br />
<input type="submit" name="add" value="Add Comment" />
</form>
However, the $assignment variable does not work in the query. It is echoed properly before the query is made but its value inside the table after the INSERT is completed is empty. What exactly is causing this?
Instead of trying to combine GET and POST, use a hidden input field:
<?php
$assignment = mysql_real_escape_string($_POST['name']); // Name is now in POST data, so swap this
echo "$assignment <br>";
if (isset($_POST['add'])) {
$user = $_POST['username'];
$text = $_POST['comment'];
$query = "INSERT INTO comments (user, text, assignment) VALUES ('$user', '$text', '$assignment')";
mysql_query($query) or die('Error, comment failed to post');
}
?>
<h1>Add Comment</h1>
<form action="log_entry.php" method="post">
<!-- Add hidden input to carry the name -->
<input type="hidden" name="name" value="<?php echo $_GET['name']; ?>"/>
<!-- Rest of the form is the same -->
Name:<br/>
<input type="text" name="username" value="" />
<br /><br />
Comment:<br />
<textarea style="height:200px;" type="text" name="comment" value="" ></textarea>
<br /><br />
<input type="submit" name="add" value="Add Comment" />
</form>
i'm trying to make CMS with PHP
i need to get the if from post request
<form action="updatevideo.php" method="post" role="form">
Title: <input name="title" type="text" required> <br />
description:<input name="desc" type="text" required> <br />
url: <input name="ytl" type="text" required> <br />
<input type="hidden" name="id" />
<input type="submit" name="addVideo" value="Add New Video" />
</form>
how can i make this input's value = id
<input type="hidden" name="id" />
on control page
public function Update()
{
/*
1-get data into variables
2-validation
3-Database
*/
if(isset($_GET['id']) && (int)$_GET['id']>0)
{
$id = (int)$_GET['id'];
$user = $this->videoModel->Get_By_Id($id);
print_r($user);
echo
'
<!DOCTYPE html>
<html>
<head>
<title>Page Title</title>
</head>
<body>
<form action="updatevideo.php" method="post" role="form">
Title: <input name="title" type="text" required> <br />
description:<input name="desc" type="text" required> <br />
url: <input name="ytl" type="text" required> <br />
<input type="hidden" name="id" />
<input type="submit" name="addVideo" value="Add New Video" />
</form>
</body>
</html>
';
}
else
{
if(isset($_POST['addVideo']))
{
$id = $_POST['id'];
echo "5ara" ;
$title = $_POST['title'];
$desc = $_POST['desc'];
$url = $_POST['ytl'];
//Validation
$vid = $this->getVid($url); //video id -_-
$data = array(
'title' => $title,
'desc' => $desc,
'vid' => $vid
);
if($this->videoModel->Update($id,$data))
{
System::Get('tpl')->assign('message','User Updated');
System::Get('tpl')->draw('success');
}
else
{
System::Get('tpl')->assign('message','Error Updating User');
System::Get('tpl')->draw('error');
}
}
else
{
System::Get('tpl')->assign('message','NO USER CHOSEN');
System::Get('tpl')->draw('error');
}
}
}
Quite simple really. As you checked it exists and moved it into a variable you can just echo the $id into the value attribute of that input tag.
<input type="hidden" name="id" value="' . $id . '" />
Using your code:
echo '
<!DOCTYPE html>
<html>
<head>
<title>Page Title</title>
</head>
<body>
<form action="updatevideo.php" method="post" role="form">
Title: <input name="title" type="text" required> <br />
description:<input name="desc" type="text" required> <br />
url: <input name="ytl" type="text" required> <br />
<input type="hidden" name="id" value="' . $id . '" />
<input type="submit" name="addVideo" value="Add New Video" />
</form>
</body>
</html>';
Define the VALUE attribute :
<input type="hidden" name="id" value="ID_VALUE"/>
OR if you have id in varible than use below code :
<input type="hidden" name="id" value="<?php echo $id; ?>"/>
OR
As you are using the HTML in PHP than:
<input type="hidden" name="id" value="'.$id.'"/>
ID_Value should be your id like 1,2,3..etc.
And Get the id on Action page :
$id = $_POST['id'];
Apart from this you have mention POST in form method and on your
action page you are trying to get the values using GET method,
which is wrong.
Edited due to question edit.
<input type="hidden" name="id" value="'.$id.'" />
You echo a big block of text, this way you can concatenate the $id by stopping block output, including $id and resuming block output.
I am encountering a strange phenomena: When I submit a form in my code, then beside doing what is there in that form, it looks like it also trigger the comment form. So when the page is reloading, I get some blank comments or duplicate comments. In my code I have several forms with submit, and one of that is the form for input comment:
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="a" value="<?php echo $_GET['id'] ?>">
<input type="hidden" name="b" value="<?php echo $_SERVER['REMOTE_ADDR'] ?>">
<input type="text" name="c" value="Name"><br>
<textarea name="d">
</textarea>
<input type="submit" />
</form>
Could any of you point out for me where it gets wrong?
As requested, I post the whole (updated) code here:
<h3>Comments</h3>
<!-- <p>Put your comments here: </p> -->
<?php
//a: commenters
$i = addslashes($_POST['a']);
$ip = addslashes($_POST['b']);
$a = addslashes($_POST['c']);
$b = addslashes($_POST['d']);
if(isset($_POST['form1'])){
if(isset($i) & isset($ip) & isset($a) & isset($b))
{
$connector= new DbConnector();
$r = mysql_query("SELECT COUNT(*) FROM `databasename`.`ban` WHERE ip=$ip"); //Check if banned
$r = mysql_fetch_array($r);
if(!$r[0]) //Phew, not banned
{ // echo "a: ".$a." b: ".$b." ip: ".$ip." i: ".$i."";
$Date4=date('Y-m-d H:i:s');
if(mysql_query("INSERT INTO `databasename`.`Comments` VALUES ('$a', '$b', '$ip', '$Date4')"))
{
?>
<script type="text/javascript">
window.location="/index.php?id=".<?php echo $i; ?>;
</script>
<?php
}
else echo "Error, in mysql query";
}
else echo "Error, You are banned.";
}
}
$x = mysql_query("SELECT * FROM `databasename`.`Comments` ORDER BY i DESC ");
while($r = mysql_fetch_object($x)) echo "<div class='c'>".$r->a."<p>".$r->b."</p> </div>";
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="a" value="<?php echo $_GET['id'] ?>">
<input type="hidden" name="b" value="<?php echo $_SERVER['REMOTE_ADDR'] ?>">
<input type="text" name="c" value="Name"><br>
<textarea name="d">
</textarea>
<input type="submit" name="form1" />
</form>
You need use isset()
in HTML submit button you need to use name attribute for each and every form,
<input type="submit" name="form1" />
IN PHP,
<?php
if(isset($_POST['form1']){
echo $_POST['a'];
}
?>
Your code,
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<input type="hidden" name="a" value="<?php echo $_GET['id']; ?>">
<input type="hidden" name="b" value="<?php echo $_SERVER['REMOTE_ADDR']; ?>">
<input type="text" name="c" value="Name"><br>
<textarea name="d"></textarea>
<input type="submit" name="form1" />
</form>
I would use:
if(!empty($_POST['form1']){
echo $_POST['a'];
}
empty() will return true if the variable is an empty string, false, array(), NULL, “0?, 0, and an unset variable.
Read more: http://www.virendrachandak.com/techtalk/php-isset-vs-empty-vs-is_null/
I'm currently in the process of coding a MySQL search page and I'm unable to work out how to make it so that if there's no data put in one form it'll do nothing, and if there's data in the other form it'll search the database for that value.
<?PHP
echo '<h3 class="hp-1">Kick Logs</h3><div class="wrapper">
<div class="logcol-1"><form name="form1" id="mainForm" method="post"enctype="multipart/form-data" action="' . $_SERVER['REQUEST_URI'] . '">
<input name="name" type="text" id="name" placeholder="Players Name">
</form>';
echo '<form name="form2" id="mainForm" method="post"enctype="multipart/form-data" action="' . $_SERVER['REQUEST_URI'] . '"><input name="reason" type="text" id="reason" placeholder="Kick Reason"></form>';
$name = mysql_real_escape_string($name);
$reason = mysql_real_escape_string($reason);
$kicklogname = mysql_query("SELECT * FROM `log1` WHERE `user` LIKE '%$name%'") or die(mysql_error());
$kicklogreason = mysql_query("SELECT * FROM `log1` WHERE `user` LIKE '%$reason%'") or die(mysql_error());
if($name == ""){
echo "You must enter a name to search"; }
else {
echo '<table width="700" border="0">
<tr class="listheader">
<td width="100" bgcolor="#afe6ff">Username</td>
<td width="220" bgcolor="#afe6ff">Reason</td>
</tr>';
while($row = mysql_fetch_array($kicklogname))
{
echo '<tr><td bgcolor="#daf4ff" class="contentleft">';
echo $row['user'];
echo '</td><td bgcolor="#eefaff" class="contentright">';
echo $row['reason'];
echo '</td></tr>';
}
echo '</table></div></div>';
}
?>
Update
Sorry, re-read the question. To check multiple forms on the same page you need to add a submit button to each form and give it a name:
<?php
if (!empty($_POST) && isset($_POST['reasonsubmit'])) {
echo 'Submitted reason form';
}
if (!empty($_POST) && isset($_POST['namesubmit'])) {
echo 'Submitted name form';
}
?>
<form action="reason.php" method="post">
<input type="text" name="reason" id="reason" />
<input type="submit" name="reasonsubmit" />
</form>
<form action="name.php" method="post">
<input type="text" name="name" id="name" />
<input type="submit" name="namesubmit" />
</form>
Alternatively, if you didn't want to give the submit button a name you can just add a hidden field to each form and you will get the same behavior.
<form action="reason.php" method="post">
<input type="hidden" name="reasonsubmit" id="reasonsubmit" />
<input type="text" name="reason" id="reason" />
<input type="submit" />
</form>
<form action="name.php" method="post">
<input type="hidden" name="namesubmit" id="namesubmit" />
<input type="text" name="name" id="name" />
<input type="submit" />
</form>
I'm trying to echo a value in a form based on number of rows returned from database query. Keep getting error Parse error: syntax error, unexpected T_ECHO, expecting ',' or ';'
As you can probably tell I'm pretty new to this. Can anyone help me echo the variable? I know that $num_rows is returning a value as using var_dump shows. Thanks
<?
if($num_rows <= 10) {
echo '</br></br><form id="h1" class="rounded" action="4.php" target=""
method="post"/>
<input type="submit" name="submit" class="button" value="10" /><br>
<input type="text" name="number_of_tests" value="'echo $num_rows;'"/>
</form>';
}
if($num_rows >10) {
echo '</br></br><form id="h2" class="rounded" action="4.php"
target="_blank" method="post"/>
<input type="submit" name="submit" class="button" value="11"/><BR>
<input type="text" name="number_of_tests" value="'echo $num_rows;'"/>
</form>';
}?>
In both of your code blocks, you repeat the command echo instead of either concatenating the output or using two statements. You have done this:
echo '</br></br><form id="h1" class="rounded" action="4.php" target=""
method="post"/>
<input type="submit" name="submit" class="button" value="10" /><br>
<input type="text" name="number_of_tests" value="'echo $num_rows;'"/>
</form>';
which is a syntax error. Instead, you can do this:
echo '</br></br><form id="h1" class="rounded" action="4.php" target=""
method="post"/>
<input type="submit" name="submit" class="button" value="10" /><br>
<input type="text" name="number_of_tests" value="' . $num_rows . '"/>
</form>';
or this:
echo '</br></br><form id="h1" class="rounded" action="4.php" target=""
method="post"/>
<input type="submit" name="submit" class="button" value="10" /><br>
<input type="text" name="number_of_tests" value="';
echo $num_rows . '"/>';
echo '</form>';
This is the code you should use to concatenate strings and output the result
echo ' some value ' . $variable . ' other text ';
The echo function outputs a string, while the dot (.) operator concatenates strings. This is the kind of wrong code
echo 'value="'echo $num_rows;'"/>';
When you want to insert the value of a variable this is the way
$a_string = 'I\'m a string';
echo "I'm a double quoted string and can contain a variable: $a_string";
This works with arrays too
$an_array = array('one', 'two', 'three');
echo "The first element of the array is {$an_array[0]}"
See the PHP manual