Add multiple rows to a single php while loop variable

Question

I currently have this code that's working just fine:

while ($row = mysqli_fetch_array($popupResult)) {
  $popupQuote     = $row['quoteNumber'];
  $popupExp       = $row['quoteCustomer'];
  $popupCost      = $row['quoteExpirationDate'];
  }

My only issue is that the data in my table that contains what I want in $popupQuote COULD have quoteNumber populated, or it could have debitNumber populated - but it will never have both. I need this basically:

  $popupQuote     = $row['quoteNumber'] or $row['debitNumber']

If I use $popupQuote = $row['quoteNumber'];, I get 6 results. If I use $popupQuote = $row['debitNumber']; I can 4 results. I want the output to display all 10 results in a single variable.

Thanks for the edit... darn copy & paste :)

Brian Powell
– Brian Powell

2014-12-18 17:02:53 +00:00
Commented Dec 18, 2014 at 17:02 — Brian Powell
– Brian Powell, Commented Dec 18, 2014 at 17:02

Marc B · Accepted Answer · 2014-12-18 17:23:48Z

3

If you guarantee that one of the two fields will always be populated, and never will there be a record with both or neither, then:

$poupQuote = !empty($row['quoteNumber']) ? $row['quoteNumber'] : $row['debitNumber'];

e.g. check if quotenumber has a value. if it does, use that value, otherwise use the debitnumber. Of course, you'll probably want some better checking, since empty() will bite you in the rump if quotenumber can be 0 - empty(0) is true.

edited Dec 18, 2014 at 17:23

answered Dec 18, 2014 at 17:01

Marc B

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1 Comment

Brian Powell Over a year ago

There needs to be a closing bracket though on ['quoteNumber']

Barmar · Accepted Answer · 2014-12-18 17:01:13Z

0

Use a conditional expression:

$popupQuote = $row['quoteNumber'] ? $row['quoteNumber'] : $row['debitNumber'];

answered Dec 18, 2014 at 17:01

Barmar

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Comments

Funk Forty Niner · Accepted Answer · 2014-12-18 17:22:06Z

0

What you need is a Ternary operator

And test if there is a value

$popupQuote = ($row['quoteNumber'] != "") ? $row['quoteNumber'] : $row['debitNumber'];

edited Dec 18, 2014 at 17:22

Funk Forty Niner

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answered Dec 18, 2014 at 17:05

Demodave

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2 Comments

Marc B Over a year ago

isset wouldn't work. the field will always be present in the result row... it just might not have a value.

Demodave Over a year ago

Oh, I see you have something similar.

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