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Script not showing a result in php if I change a condition in sql.

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Apr 27.2015

I am using[B] mssql_query[/B]. I have this sql query that have [B]joins[/B] and [B]distinct[/B]. If I change the condition value it will not show a result when run as a php script. But if I run that query directly in a sql server browser, it will show results.

For example below, if I change the condition [B]SOTYPE = ‘aaa’[/B] to [B]SOTYPE = ‘bbb’[/B],it will not show any result if it will be run as php script.But using directly the sql server browser, it will display results.

Now if the condition is [B]SOTYPE = ‘aaa’[/B] it will also show results,run as php script. Now removing [B]distinct[/B] it will show results(run as php script) no matter what conditions that I do.

Is this php memory related? How can I solve this problem?

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6 Comments(s) ↴

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@r8rApr 27.2015 — This may be due to escaping issues and maybe you need to use GROUP BY also. However, without actually seeing a little more of your code and queries, I'm just stabbing in the dark.
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@anishgiriauthorApr 27.2015 — [i][noparse][Edited to replace tags with and [code=php] tags as appliable. ~ Mod.][/noparse][/i]

This may be due to escaping issues and maybe you need to use GROUP BY also. However, without actually seeing a little more of your code and queries, I'm just stabbing in the dark.[/QUOTE]

My php code is just composed of this query and trying to see if it will show results.
[code]
SELECT DISTINCT prd.ZSD_OMSH.*, prd.KNA1.NAME1 AS K_NAME1, prd.KNA1.NAME2 AS K_NAME2, prd.ZSD_OMSD.WERKS, prd.ZSD_OMSD.LGORT, prd.KNA1.NAME3 AS K_NAME3, prd.KNA1.NAME4 AS K_NAME4, prd.PA0001.ENAME FROM prd.ZSD_OMSH INNER JOIN prd.KNA1 ON prd.ZSD_OMSH.KUNNR = prd.KNA1.KUNNR INNER JOIN prd.PA0001 ON prd.ZSD_OMSH.PERNR = prd.PA0001.PERNR INNER JOIN prd.ZSD_OMSD ON prd.ZSD_OMSH.ORDNUM = prd.ZSD_OMSD.ORDNUM where prd.ZSD_OMSH.ERDAT >= '20150301' and prd.ZSD_OMSH.ERDAT <= '20150424' and prd.ZSD_OMSH.MANDT = '888' and prd.KNA1.MANDT = '888' and prd.PA0001.MANDT = '888' and SOTYPE = 'ZKB' and ((DISAPPROVE is null) or (DISAPPROVE <> 1)) and ((VSTATIMD2 is null) or (VSTATIMD2 IN('0',''))) and NEXT_APP = '4'
[code=php]
$rs = mssql_query($sql);
if (mssql_num_rows($rs) > 0) {
while ($row = mssql_fetch_array($rs)) {

echo "result";
}
}[/code]

This sql query above will show result if I directly use sql server browser. But if I run it as a php script it will not show result. But if i change the condition [B]SOTYPE = 'ZKB' to SOTYPE = 'ZOR'[/B], it will show result on my php script.
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@ginerjmApr 27.2015 — Have you checked your error log for any messages? Or better yet, have you enabled error checking so that error messages will display onscreen for you? (See my signature.)
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@NogDogApr 27.2015 — For now, I would double-check that the query being submitted is actually what you think it is:
[code=php]
$rs = mssql_query($sql);
if (mssql_num_rows($rs) > 0) {
while ($row = mssql_fetch_array($rs)) {

echo "result";
}
}
else {
echo "<pre>DEBUG (num rows 0):n$sql</pre>";
}
[/code]
Then you can copy/paste the SQL exactly as it was at that point in the script and see what you get.
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@anishgiriauthorApr 28.2015 — For now, I would double-check that the query being submitted is actually what you think it is:
[code=php]
$rs = mssql_query($sql);
if (mssql_num_rows($rs) > 0) {
while ($row = mssql_fetch_array($rs)) {

echo "result";
}
}
else {
echo "<pre>DEBUG (num rows 0):n$sql</pre>";
}
[/code]
Then you can copy/paste the SQL exactly as it was at that point in the script and see what you get.[/QUOTE]

Yeah I think it's the actual query that is being submitted because actualy the sql query that I posted here is the one that I [B]echo/print[/B]. Turning on error message log it says.

Warning: mssql_fetch_array(): supplied argument is not a valid MS SQL-result resource in c:inetpubwwwrootermsapoms_approval_report_sap.php on line 334.

But just now, the script is already showing result/data. And I did not modify any code. There are many people trying to access this data a while ago. Maybe the data that is being retrieve is just to huge and this cause this sql error?
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@NogDogApr 28.2015 — I suppose there could me several potential reason, which is why it's always a good idea to check the return value of a call to *_query() and make sure it didn't choke on something. I'm not conversant with mssql_*() functions, but it looks like you could use mssql_get_last_message() to find out what happened, as in the example on the manual page: http://php.net/manual/en/function.mssql-get-last-message.php . (In the ultimate production code, you probably don't want to die() and spew debug info, but instead error_log() the info and display some sort of "nice" error page. ? )
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