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Closed 9 years ago.
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Hello i'm new in PHP and i'm trying to upload 3 images and i got an error
Please don't downgrade me i'm just learning and trying to make a simple system. Thank you
Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\epushmoyan\customer\connect.php on line 17
Here's my code :
function addpost($a,$b,$c,$d,$name,$type,$size,$tmp_name,$error,$name1,$type1,$size1,$tmp_name1,$error1,$name2,$type2,$size2,$tmp_name2,$error2,$e)
{
$img = 'thumbnails/products/'.$name.$name1.$name2
mysql_query("INSERT post(p_name,p_contact,p_email,image,image1,image2,p_address,dateadded) values ('$a','$b','$c','$d','$img','$e',NOW())");
if ($error > 0)
{
die("Error uploading file! Code $error.");
}
else
{
if($size> 10000000000) //conditions for the file
{
die("Format is not allowed or file size is too big!");
}
else
{
move_uploaded_file($tmp_name,$tmp_name1,$tmp_name2,"thumbnails/products/".$name.$name1,$name2);
}
}
Missing ;:
$img = 'thumbnails/products/'.$name.$name1.$name2
^--- here
mysql_query
Voting to close, since it's a simple typo.
Missing ; in
$img = 'thumbnails/products/'.$name.$name1.$name2
Related
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Closed 9 months ago.
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I want to enter an multiple field entered data in table with for loop
but I am getting an error in the post method.
The error is:
Use of undefined constant i - assumed 'i' (this will throw an Error in a future version of PHP)
controller code:
$degree = $request->degree;
if($degree > 0)
{
for($i=0;$i<count($degree);$i++){
$edu = new education;
$edu->degree = $request->degree[i];
$edu->clg = $request->clg[i];
$edu->yoc = $request->yoc[i];
$edu->save();
}
}
so, please suggest me what I can do.
here there is a silly mistake bro you not remember to use $i inside loop for the degree, clg and yoc
$degree = $request->degree;
if($degree > 0)
{
for($i=0;$i<count($degree);$i++){
$edu = new education;
$edu->degree = $request->degree[$i];
$edu->clg = $request->clg[$i];
$edu->yoc = $request->yoc[$i];
$edu->save();
}
}
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Closed 7 years ago.
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I have two PHP scripts, dispatcher.php and processor.php (there are more, but I've commented the others out because they're not necessary at the moment).
I get the following error:
Parse error: syntax error, unexpected '$this' (T_VARIABLE) in /----home directory-----/classes/processor.php on line 12
Here's the code for processor.php:
class Processor{
protected $player;
protected $name;
protected $id;
function __construct(){
}
function loadvars($request){
$this->loadvar($this->name, $request, "name");
}
private function loadvar($target, $request, $name){
if(isset($request[$name])){
$target = $request[$name];
}
else{
$target = "";
}
}
}
this is the code for dispatcher.php:
require('classes/processor.php');
$test = new Processor();
$test->loadvars($_GET);
I don't see why this error occurs?
I'm coding using Sublime on Windows, if that matters.
Fixed it!
Try retyping the file if you have weird parse errors. I copied the code back from StackOverflow (where I typed it above) and it worked. Guess some unparseable character got into the file somewhere.
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
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Closed 8 years ago.
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I have been trying to put up a small testing site for a project at codebase.host22.org, and I cant seem to get my PHP script to work. The script is hown as follows:
<?php
$browser = get_browser(null, true);
$data = $browser[browser]
$f = fopen("data.csv", "a");
fwrite($f, $data = ",");
fclose($f);
?>
I don't see a problem. But I get this error when trying to visit the page:
Parse error: syntax error, unexpected T_VARIABLE in /home/a5547326/public_html/index.php on line 6
My hosting is with 000webhost. Is there a problem with my code or the server?
You missed semicolon after $data variable.
$data = $browser[browser];
You're missing the quote and the semicolon:
$data = $browser['browser'];
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
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It's a basic question, but I've googled all sorts of variations of it & nothing that quite answers me. I want to display one of two images. One if the criteria is met & one if it is not met. This code works. If criteria is met, it does display the image. But there is no alternative
<? if(stripslashes($getfeedbackQryRow['CompanyID'])=='344'){?><img src='img1.png' alt='Todays Bite'><? }?>
Then I put this together, but it's still wrong
<?PHP
if ($getfeedbackQryRow['CompanyID']) == '344') {
print ("<IMG SRC =/img1.png>");
}
else {
print ("<IMG SRC =img2.png>");
}
?>
This one looks like it should be right, but it throws an error... It's probably a syntax issue. Does anyone know what I'm doing wrong?
You have an extra ")"
if ($getfeedbackQryRow['CompanyID']) == '344') {
Remove the ")" here ['CompanyID'])
if ($getfeedbackQryRow['CompanyID']) == '344') {
// ^------- this is causing your error
You should do something simple like this:
$image_name = $getfeedbackQryRow['CompanyId'] == '344' ? 'img1.png' : 'img2.png';
echo '<img src="' . $image_name . '">';
Closed. This question is not reproducible or was caused by typos. It is not currently accepting answers.
This question was caused by a typo or a problem that can no longer be reproduced. While similar questions may be on-topic here, this one was resolved in a way less likely to help future readers.
Closed 8 years ago.
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I am trying to write a program using Call By Reference to calculate the area but I am getting a Parse Error :-
Parse error: syntax error, unexpected 'ar' (T_STRING) in C:\xampp\htdocs\workspace.php on line 7
Now I, cannot think why it is happening?
<?php
function perimeter(&$l,&$b,&$result)
{
$result=2*($l+$b);
}
funtction ar(&$le,&$br,&$result1)
{
$result1=$le*$br;
}
$result=1;
$length=$_GET['length'];
$breadth=$_GET['breadth'];
echo "<h1><center>Area And Perimeter Calculator Using Call By Reference</h1></center>";
$result = perimeter($length,$breadth,$result);
echo "<br />The Perimeter Of The Rectangle Is <strong>$result</strong>";
$result= ar($length,$breadth,$result);
echo "<br /><br />The Area Of The Rectangle Is = <strong>$result</strong>";
?>
You misspelled function:
funtction ar(&$le,&$br,&$result1)
{
$result1=$le*$br;
}
should be
function ar(&$le,&$br,&$result1)
{
$result1=$le*$br;
}