Re: [PHP-DB] weird string literal related problem

[Ratin]

nvm, I got my answer -- foreach ( $entries as $column => $value ) will give
the column names .. which simplifies my task a lot.

On Mon, Aug 1, 2016 at 12:52 PM, Ratin  wrote:

> Just curious why the values can only be accessed by column names and not
> an array subscript, like :
>
> $i=0;
> $result = $db->query('select * from DecoderUser');
> while ($entries = $result->fetchArray())
> {
> $value=$entries[i];
> $i = $i + 1;
> }
>
> On Mon, Aug 1, 2016 at 12:25 PM, Ratin  wrote:
>
>> Thanks Aziz, that works. Somehow i thought the ' needed to be included
>> just like you quote all strings on an interactive DB shell or even when you
>> get the value directly by column name.
>>
>> On Mon, Aug 1, 2016 at 11:47 AM, Aziz Saleh  wrote:
>>
>>> But if i print $entries['decoderUserGUID'] I get the correct value
>>>
>>> This means the key value is decoderUserGUID NOT 'decoderUserGUID', try
>>> creating $strIndex without the quotes, just the key value.
>>>
>>> On Mon, Aug 1, 2016 at 2:31 PM, Ratin  wrote:
>>>
 Hi I am new to this list, not sure how active this list is but i have a
 question regarding php-sqlite integration. I am using a string variable
 where I put the column name and trying to use that varible to get its
 value, in a loop. So here is what I have (simplified version):


 $strIndex='\'' . $columns[$i] . '\'';  //strIndex has the value :
 'decoderUserGUID'  (including quotes)
 $result = $db->query('select * from DecoderUser');
 while ($entries = $result->fetchArray())
 {
 $value=$entries[$strIndex];
 }



 So if I print $value, I get nothing. But if i print
 $entries['decoderUserGUID'] I get the correct value. If I print
 $strIndex,
 I get 'decoderUserGUID' (including the quotes). How can I solve this
 mistery.. whats hapenning here , Anybody have any idea?


 Thanks

 Ratin

>>>
>>>
>>
>

Re: [PHP-DB] weird string literal related problem

[Ratin]

Just curious why the values can only be accessed by column names and not an
array subscript, like :

$i=0;
$result = $db->query('select * from DecoderUser');
while ($entries = $result->fetchArray())
{
$value=$entries[i];
$i = $i + 1;
}

On Mon, Aug 1, 2016 at 12:25 PM, Ratin  wrote:

> Thanks Aziz, that works. Somehow i thought the ' needed to be included
> just like you quote all strings on an interactive DB shell or even when you
> get the value directly by column name.
>
> On Mon, Aug 1, 2016 at 11:47 AM, Aziz Saleh  wrote:
>
>> But if i print $entries['decoderUserGUID'] I get the correct value
>>
>> This means the key value is decoderUserGUID NOT 'decoderUserGUID', try
>> creating $strIndex without the quotes, just the key value.
>>
>> On Mon, Aug 1, 2016 at 2:31 PM, Ratin  wrote:
>>
>>> Hi I am new to this list, not sure how active this list is but i have a
>>> question regarding php-sqlite integration. I am using a string variable
>>> where I put the column name and trying to use that varible to get its
>>> value, in a loop. So here is what I have (simplified version):
>>>
>>>
>>> $strIndex='\'' . $columns[$i] . '\'';  //strIndex has the value :
>>> 'decoderUserGUID'  (including quotes)
>>> $result = $db->query('select * from DecoderUser');
>>> while ($entries = $result->fetchArray())
>>> {
>>> $value=$entries[$strIndex];
>>> }
>>>
>>>
>>>
>>> So if I print $value, I get nothing. But if i print
>>> $entries['decoderUserGUID'] I get the correct value. If I print
>>> $strIndex,
>>> I get 'decoderUserGUID' (including the quotes). How can I solve this
>>> mistery.. whats hapenning here , Anybody have any idea?
>>>
>>>
>>> Thanks
>>>
>>> Ratin
>>>
>>
>>
>

Re: [PHP-DB] weird string literal related problem

[Ratin]

Thanks Aziz, that works. Somehow i thought the ' needed to be included just
like you quote all strings on an interactive DB shell or even when you get
the value directly by column name.

On Mon, Aug 1, 2016 at 11:47 AM, Aziz Saleh  wrote:

> But if i print $entries['decoderUserGUID'] I get the correct value
>
> This means the key value is decoderUserGUID NOT 'decoderUserGUID', try
> creating $strIndex without the quotes, just the key value.
>
> On Mon, Aug 1, 2016 at 2:31 PM, Ratin  wrote:
>
>> Hi I am new to this list, not sure how active this list is but i have a
>> question regarding php-sqlite integration. I am using a string variable
>> where I put the column name and trying to use that varible to get its
>> value, in a loop. So here is what I have (simplified version):
>>
>>
>> $strIndex='\'' . $columns[$i] . '\'';  //strIndex has the value :
>> 'decoderUserGUID'  (including quotes)
>> $result = $db->query('select * from DecoderUser');
>> while ($entries = $result->fetchArray())
>> {
>> $value=$entries[$strIndex];
>> }
>>
>>
>>
>> So if I print $value, I get nothing. But if i print
>> $entries['decoderUserGUID'] I get the correct value. If I print $strIndex,
>> I get 'decoderUserGUID' (including the quotes). How can I solve this
>> mistery.. whats hapenning here , Anybody have any idea?
>>
>>
>> Thanks
>>
>> Ratin
>>
>
>

Re: [PHP-DB] weird string literal related problem

[Aziz Saleh]

But if i print $entries['decoderUserGUID'] I get the correct value

This means the key value is decoderUserGUID NOT 'decoderUserGUID', try
creating $strIndex without the quotes, just the key value.

On Mon, Aug 1, 2016 at 2:31 PM, Ratin  wrote:

> Hi I am new to this list, not sure how active this list is but i have a
> question regarding php-sqlite integration. I am using a string variable
> where I put the column name and trying to use that varible to get its
> value, in a loop. So here is what I have (simplified version):
>
>
> $strIndex='\'' . $columns[$i] . '\'';  //strIndex has the value :
> 'decoderUserGUID'  (including quotes)
> $result = $db->query('select * from DecoderUser');
> while ($entries = $result->fetchArray())
> {
> $value=$entries[$strIndex];
> }
>
>
>
> So if I print $value, I get nothing. But if i print
> $entries['decoderUserGUID'] I get the correct value. If I print $strIndex,
> I get 'decoderUserGUID' (including the quotes). How can I solve this
> mistery.. whats hapenning here , Anybody have any idea?
>
>
> Thanks
>
> Ratin
>