Re: [PHP] fgetcsv doesn't return an array?
Tamara Temple wrote:
On Thu, 15 Mar 2012 12:09:53 -0500, Jay Blanchard
sent:
I thought that fgetcsv returned an array. I can work with it like an
array but I get the following warning when using it
|Warning: implode(): Invalid arguments passed on line 155
154$csvCurrentLine = fgetcsv($csvFile, 4096, ',');
155$currentLine = implode(",", $csvCurrentLine);
|
Everything that I have read online indicates that it is because
$csvCurrentLine is not declared as an array. Adding a line to the code
153 |$csvCurrentLine = array();
Does not get rid of the warning. The warning isn't interfering with the
script working, but it us annoying. Does anyone know the solution?
Just want to clarify something not directly related to the OP's question:
This:
153 |$csvCurrentLine = array();
Does not set a *type* for $csvCurrentLine, so if you're next line is this:
154$csvCurrentLine = fgetcsv($csvFile, 4096, ',');
Any previous assignments are lost; the notion that presetting $csvCurrentLine to
an array has no bearing when you then immediately set the same variable to
something else.
The "type" $csvCurrentLine has after line 154 is entirely dependent on what
fgetcsv() is returning, and nothing that happened prior to line 154.
In this case, Jay's attempt to fix the problem was simply wrong. fgetcsv may
well not return an array, and the error in this code was simply that situation.
If an array is not returned by fgetcsv then there was an error, hence line 155
fails every time you hit the 'EOF' condition.
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Re: [PHP] fgetcsv doesn't return an array?
On Thu, 15 Mar 2012 12:09:53 -0500, Jay Blanchard
sent:
I thought that fgetcsv returned an array. I can work with it like an
array but I get the following warning when using it
|Warning: implode(): Invalid arguments passed on line 155
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
155 $currentLine = implode(",", $csvCurrentLine);
|
Everything that I have read online indicates that it is because
$csvCurrentLine is not declared as an array. Adding a line to the code
153 |$csvCurrentLine = array();
Does not get rid of the warning. The warning isn't interfering with the
script working, but it us annoying. Does anyone know the solution?
Thanks!
Jay
|
Just want to clarify something not directly related to the OP's question:
This:
153 |$csvCurrentLine = array();
Does not set a *type* for $csvCurrentLine, so if you're next line is this:
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
Any previous assignments are lost; the notion that presetting
$csvCurrentLine to an array has no bearing when you then immediately
set the same variable to something else.
The "type" $csvCurrentLine has after line 154 is entirely dependent on
what fgetcsv() is returning, and nothing that happened prior to line
154.
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aka tamouse__
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Re: [PHP] fgetcsv doesn't return an array?
[snip] Not sure why, but ok. Thanks everyone! An empty line would lead to $csvCurrentLine == array() which as a boolean would be false but would not necessarily mean EOF. By using === to check the value the expression will only evaluate to false at EOF or an error. -Stuart [/snip] Makes perfect sense! I just wasn't reading it right. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] fgetcsv doesn't return an array?
On 15 Mar 2012, at 20:26, Jay Blanchard wrote:
> [snip]
>> Please, don't do that. Use this instead:
>> while (($csvCurrentLine = fgetcsv($csvFile, 4096, ',')) !== FALSE) {
>> }
>> [/snip]
>
> Not sure why, but ok. Thanks everyone!
An empty line would lead to $csvCurrentLine == array() which as a boolean would
be false but would not necessarily mean EOF. By using === to check the value
the expression will only evaluate to false at EOF or an error.
-Stuart
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Re: [PHP] fgetcsv doesn't return an array?
[snip]
> Please, don't do that. Use this instead:
> while (($csvCurrentLine = fgetcsv($csvFile, 4096, ',')) !== FALSE) {
> }
> [/snip]
Not sure why, but ok. Thanks everyone!
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Re: [PHP] fgetcsv doesn't return an array?
On Thu, Mar 15, 2012 at 6:40 PM, Andrew Ballard wrote:
> On Thu, Mar 15, 2012 at 1:29 PM, Jay Blanchard
> wrote:
>> On 3/15/2012 12:23 PM, Matijn Woudt wrote:
>>>
>>> On Thu, Mar 15, 2012 at 6:09 PM, Jay Blanchard
>>> wrote:
I thought that fgetcsv returned an array. I can work with it like an
array
but I get the following warning when using it
|Warning: implode(): Invalid arguments passed on line 155
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
155 $currentLine = implode(",", $csvCurrentLine);
|
Everything that I have read online indicates that it is because
$csvCurrentLine is not declared as an array. Adding a line to the code
153 |$csvCurrentLine = array();
Does not get rid of the warning. The warning isn't interfering with the
script working, but it us annoying. Does anyone know the solution?
Thanks!
Jay
|
>>>
>>> Are you using this in a loop or something? fgetcsv returns FALSE on
>>> errors, including End Of File.
>>>
>>> [/snip]
>>
>> I am using it in a loop. End Of File is an error?
>
> Yes. "fgetcsv() returns NULL if an invalid handle is supplied or
> FALSE on other errors, including end of file."
>
> I usually use it in a while loop like this:
>
>
> while ($row = fgetcsv($csvFilePointer)) {
> // ... process the row
> }
>
> ?>
Please, don't do that. Use this instead:
while (($csvCurrentLine = fgetcsv($csvFile, 4096, ',')) !== FALSE) {
}
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Re: [PHP] fgetcsv doesn't return an array?
Jay Blanchard wrote:
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
155 $currentLine = implode(",", $csvCurrentLine);
I am using it in a loop. End Of File is an error?
You certainly need to break out on $csvCurrentLine = false which indicates end
of file at least.
My own 'loop' is
$row=0;
while (($data = fgetcsv($handle, 800, ",")) !== FALSE) {
if ( $row ) $contact->RecordLoad( $data, $row );
$row++;
}
The row bit strips the header line from the csv file ...
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-
Contact - http://lsces.co.uk/wiki/?page=contact
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Re: [PHP] fgetcsv doesn't return an array?
On Thu, Mar 15, 2012 at 1:29 PM, Jay Blanchard
wrote:
> On 3/15/2012 12:23 PM, Matijn Woudt wrote:
>>
>> On Thu, Mar 15, 2012 at 6:09 PM, Jay Blanchard
>> wrote:
>>>
>>> I thought that fgetcsv returned an array. I can work with it like an
>>> array
>>> but I get the following warning when using it
>>>
>>> |Warning: implode(): Invalid arguments passed on line 155
>>>
>>> 154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
>>> 155 $currentLine = implode(",", $csvCurrentLine);
>>> |
>>>
>>> Everything that I have read online indicates that it is because
>>> $csvCurrentLine is not declared as an array. Adding a line to the code
>>>
>>> 153 |$csvCurrentLine = array();
>>>
>>> Does not get rid of the warning. The warning isn't interfering with the
>>> script working, but it us annoying. Does anyone know the solution?
>>>
>>> Thanks!
>>>
>>> Jay
>>> |
>>
>> Are you using this in a loop or something? fgetcsv returns FALSE on
>> errors, including End Of File.
>>
>> [/snip]
>
> I am using it in a loop. End Of File is an error?
Yes. "fgetcsv() returns NULL if an invalid handle is supplied or
FALSE on other errors, including end of file."
I usually use it in a while loop like this:
Andrew
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Re: [PHP] fgetcsv doesn't return an array?
On 03/15/2012 10:09 AM, Jay Blanchard wrote:
I thought that fgetcsv returned an array. I can work with it like an
array but I get the following warning when using it
|Warning: implode(): Invalid arguments passed on line 155
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
155 $currentLine = implode(",", $csvCurrentLine);
|
Everything that I have read online indicates that it is because
$csvCurrentLine is not declared as an array. Adding a line to the code
153 |$csvCurrentLine = array();
Does not get rid of the warning. The warning isn't interfering with the
script working, but it us annoying. Does anyone know the solution?
Thanks!
Jay
|
From the manual
"fgetcsv() returns NULL if an invalid handle is supplied or FALSE on
other errors, including end of file."
do this
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
var_dump($csvCurrentLine);
155 $currentLine = implode(",", $csvCurrentLine);
What does it say about the variable from the failing line?
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Re: [PHP] fgetcsv doesn't return an array?
On 3/15/2012 12:23 PM, Matijn Woudt wrote:
On Thu, Mar 15, 2012 at 6:09 PM, Jay Blanchard
wrote:
I thought that fgetcsv returned an array. I can work with it like an array
but I get the following warning when using it
|Warning: implode(): Invalid arguments passed on line 155
154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
155 $currentLine = implode(",", $csvCurrentLine);
|
Everything that I have read online indicates that it is because
$csvCurrentLine is not declared as an array. Adding a line to the code
153 |$csvCurrentLine = array();
Does not get rid of the warning. The warning isn't interfering with the
script working, but it us annoying. Does anyone know the solution?
Thanks!
Jay
|
Are you using this in a loop or something? fgetcsv returns FALSE on
errors, including End Of File.
[/snip]
I am using it in a loop. End Of File is an error?
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Re: [PHP] fgetcsv doesn't return an array?
On Thu, Mar 15, 2012 at 6:09 PM, Jay Blanchard
wrote:
> I thought that fgetcsv returned an array. I can work with it like an array
> but I get the following warning when using it
>
> |Warning: implode(): Invalid arguments passed on line 155
>
> 154 $csvCurrentLine = fgetcsv($csvFile, 4096, ',');
> 155 $currentLine = implode(",", $csvCurrentLine);
> |
>
> Everything that I have read online indicates that it is because
> $csvCurrentLine is not declared as an array. Adding a line to the code
>
> 153 |$csvCurrentLine = array();
>
> Does not get rid of the warning. The warning isn't interfering with the
> script working, but it us annoying. Does anyone know the solution?
>
> Thanks!
>
> Jay
> |
Are you using this in a loop or something? fgetcsv returns FALSE on
errors, including End Of File.
- Matijn
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