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Keeps telling me Every derived table must have its own alias for the following:

[code]
SELECT zip_code.* FROM zip_code LEFT JOIN zip_tracker ON zip_code.zip_code_id=zip_tracker.zip_code_id,

(SELECT SUM(zc.quantity) AS total FROM zip_code as zc LEFT JOIN zip_tracker zt ON zc.zip_code_id=zt.zip_code_id WHERE zt.order_id='50')

WHERE zip_tracker.order_id='50'
[/code]
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https://forums.phpfreaks.com/topic/3312-what-is-wrong-with-this/
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[!--quoteo(post=342403:date=Feb 3 2006, 09:26 AM:name=tobeyt23)--][div class=\'quotetop\']QUOTE(tobeyt23 @ Feb 3 2006, 09:26 AM) [snapback]342403[/snapback][/div][div class=\'quotemain\'][!--quotec--]
Keeps telling me Every derived table must have its own alias for the following:

[code]
SELECT zip_code.* FROM zip_code LEFT JOIN zip_tracker ON zip_code.zip_code_id=zip_tracker.zip_code_id,

(SELECT SUM(zc.quantity) AS total FROM zip_code as zc LEFT JOIN zip_tracker zt ON zc.zip_code_id=zt.zip_code_id WHERE zt.order_id='50')

WHERE zip_tracker.order_id='50'
[/code]
[/quote]

should be

[code]
SELECT zip_code.* FROM zip_code LEFT JOIN zip_tracker ON zip_code.zip_code_id=zip_tracker.zip_code_id,

(SELECT SUM(zc.quantity) AS total FROM zip_code as zc LEFT JOIN zip_tracker AS zt ON zc.zip_code_id=zt.zip_code_id WHERE zt.order_id='50')

WHERE zip_tracker.order_id='50'
[/code]


looks like you just forgot an "AS" for zt
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https://forums.phpfreaks.com/topic/3312-what-is-wrong-with-this/#findComment-11293
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No, that's not it -- the AS is optional. However, you _do_ need to give your SUM sub-query a column alias, e.g.:

[code]SELECT zip_code.* FROM zip_code LEFT JOIN zip_tracker ON zip_code.zip_code_id=zip_tracker.zip_code_id,

(SELECT SUM(zc.quantity) AS total FROM zip_code as zc LEFT JOIN zip_tracker zt ON zc.zip_code_id=zt.zip_code_id WHERE zt.order_id='50') AS sumQuantity

WHERE zip_tracker.order_id='50'[/code]

Hope that helps.
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https://forums.phpfreaks.com/topic/3312-what-is-wrong-with-this/#findComment-11294
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