[PHP] Output Blank?
Having some issues with outputting my table data as an array.
In the code below I am outputting the column titles of my table into an
excel spreadsheet. I get the column titles just fine in Excel.
if($numberFields) { // Check if we need to output anything
$types = ifx_fieldtypes($query);
if (isset($types)) {
foreach($types as $field_name[] = $data_type) {
}
}
$headers = join(',', $field_name).\n; // Make our first row in the CSV
After that I am pulling all of the column data to place under it's
respective title.
Uncommenting the print_r($info); below does display all of the data for each
column correctly. However, after I run my foreach loop and output the report
to Excel all I get is empty rows/columns where the data should be...on the
bright side I get the exact number of empty rows that my query should
return.
Any ideas why the data isn't populating? If I change $row[] =
parseCSVComments($info-$fieldName); to $row[] = parseCSVComments($info); I
get Array printed out in every cell.
while($info = ifx_fetch_row($query)) {
//print_r($info);
foreach($field_name as $fieldName) { // Loop through the array of headers
as we fetch the data
$row[] = parseCSVComments($info-$fieldName);
} // End loop
$data .= join(',', $row).\n; // Create a new row of data and append it
to the last row
$row = ''; // Clear the contents of the $row variable to start a new row
}
// Start our output of the CSV
header(Content-type: application/x-msdownload);
header(Content-Disposition: attachment; filename=data.csv);
header(Pragma: no-cache);
header(Expires: 0);
echo $headers.$data;
}
Thanks,
Dan
Re: [PHP] Output Blank?
On Mon, Jun 23, 2008 at 1:46 PM, Dan Shirah [EMAIL PROTECTED] wrote:
if($numberFields) { // Check if we need to output anything
$types = ifx_fieldtypes($query);
if (isset($types)) {
foreach($types as $field_name[] = $data_type) {
}
}
Is there a particular reason you're telling PHP that, for each
$types variable, set this key of the $field_name array to equal
$data_type, and do nothing else about it?
Is this what you meant to do for some reason?
?php
if(isset($types)) {
foreach($types as $field_name = $data_type) {
$$field_name = $data_type;
// Or should it be something like this?
$column_names[] = $field_name;
}
}
$headers = join(',', $column_names).\n;
?
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Re: [PHP] Output Blank?
When I have = $data_type in there my column title values for $field_name
display correctly. If I take out = $data_type it displays the the column
type. IE SQLNUMINT SQLCHAR...instead of the column title.
On 6/23/08, Daniel Brown [EMAIL PROTECTED] wrote:
On Mon, Jun 23, 2008 at 1:46 PM, Dan Shirah [EMAIL PROTECTED] wrote:
if($numberFields) { // Check if we need to output anything
$types = ifx_fieldtypes($query);
if (isset($types)) {
foreach($types as $field_name[] = $data_type) {
}
}
Is there a particular reason you're telling PHP that, for each
$types variable, set this key of the $field_name array to equal
$data_type, and do nothing else about it?
Is this what you meant to do for some reason?
?php
if(isset($types)) {
foreach($types as $field_name = $data_type) {
$$field_name = $data_type;
// Or should it be something like this?
$column_names[] = $field_name;
}
}
$headers = join(',', $column_names).\n;
?
--
/Daniel P. Brown
Dedicated Servers - Intel 2.4GHz w/2TB bandwidth/mo. starting at just
$59.99/mo. with no contract!
Dedicated servers, VPS, and hosting from $2.50/mo.
Re: [PHP] Output Blank?
Is there a particular reason you're telling PHP that, for each
$types variable, set this key of the $field_name array to equal
$data_type, and do nothing else about it?
The ifx_fieldtypes function returns the data as an array with the field
names as the key, and the field types as the data. So, to get the column
titles to display, I am setting creating an array for the field names(key)
by passing the =. I think?
I could rename $data_type to $apples_and_oranges or any variable name I
wanted. It just needs the reference so it knows to output the field name
instead of the field type. If I'm understanding the manual. Yes, Dan...I
read the manual this time! :)
I believe the problem I am having is within the code below:
while($info = ifx_fetch_row($query)) {
//print_r($info);
foreach($field_name as $fieldName) { // Loop through the array of headers
as we fetch the data
$row[] = parseCSVComments($info-$fieldName);
} // End loop
The output of print_r($info); is everything that I want to see and it is
actually there. But when I try to get that data into an array to display in
my table, it isn't being displayed. I think I might have the syntax
incorrect for $row[] because if I type in $row[] =
parseCSVComments($info); my output displays Array in every cell.
Dan
Re: [PHP] Output Blank?
Dan Shirah wrote:
Having some issues with outputting my table data as an array.
In the code below I am outputting the column titles of my table into an
excel spreadsheet. I get the column titles just fine in Excel.
if($numberFields) { // Check if we need to output anything
$types = ifx_fieldtypes($query);
if (isset($types)) {
foreach($types as $field_name[] = $data_type) {
}
}
$headers = join(',', $field_name).\n; // Make our first row in the CSV
After that I am pulling all of the column data to place under it's
respective title.
Uncommenting the print_r($info); below does display all of the data for each
column correctly. However, after I run my foreach loop and output the report
to Excel all I get is empty rows/columns where the data should be...on the
bright side I get the exact number of empty rows that my query should
return.
Any ideas why the data isn't populating? If I change $row[] =
parseCSVComments($info-$fieldName); to $row[] = parseCSVComments($info); I
get Array printed out in every cell.
while($info = ifx_fetch_row($query)) {
//print_r($info);
foreach($field_name as $fieldName) { // Loop through the array of headers
as we fetch the data
$row[] = parseCSVComments($info-$fieldName);
} // End loop
$data .= join(',', $row).\n; // Create a new row of data and append it
to the last row
$row = ''; // Clear the contents of the $row variable to start a new row
}
// Start our output of the CSV
header(Content-type: application/x-msdownload);
header(Content-Disposition: attachment; filename=data.csv);
header(Pragma: no-cache);
header(Expires: 0);
echo $headers.$data;
}
Thanks,
Dan
?php
while($info = ifx_fetch_row($query)) {
//print_r($info);
// Initialize/clear the contents of the $row variable
$row = array();
// You probably need to reset the $field_name var with each
// iteration over it. from what I understand, foreach will move
// the internal pointer of the array to the end. Then the next
// iteration over the array will start at the end. BAD!!!
reset($field_name);
// Loop through the array of headers as we fetch the data
foreach($field_name as $fieldName) {
// From the manual: ifx_fetch_row returns an associative array
$row[] = parseCSVComments($info[$fieldName]);
} // End loop
// Create a new row of data and append it to the last row
$data .= join(',', $row).\n;
} // End while
?
BTW: what are you doing, if anything, if your data has a comma in it?
--
Jim Lucas
Some men are born to greatness, some achieve greatness,
and some have greatness thrust upon them.
Twelfth Night, Act II, Scene V
by William Shakespeare
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Re: [PHP] Output Blank?
AH HA! Got it!
The problem was I did not need to loop through the array headers to
associate the data. So I removed:
foreach($field_name as $fieldName) { // Loop through the array of headers
as we fetch the data
$row[] = parseCSVComments($info-$fieldName);
} // End loop
And ran it as:
while($info = ifx_fetch_row($query)) {
//print_r($info);
$data .= join(',', $row).\n; // Create a new row of data and append it
to the last row
$row = ''; // Clear the contents of the $row variable to start a new row
}
And everything output correctly!
Dan
Re: [PHP] Output Blank?
// Initialize/clear the contents of the $row variable $row = array(); // You probably need to reset the $field_name var with each // iteration over it. from what I understand, foreach will move // the internal pointer of the array to the end. Then the next // iteration over the array will start at the end. BAD!!! reset($field_name); BTW: what are you doing, if anything, if your data has a comma in it? Jim, I took your advice about clearing the array and resetting the $field_name variable. Thanks! I am not worried about comma's in my data because my application and database do not allow comma's to be input for this information. :) Dan
