How do I apply results after a multi-table query?

Source: Internet
Author: User
Title.. I can only have enough of the contents of a table, which is why? My SQL statement is correct.


Reply to discussion (solution)

... ... If you ask a concise question next time, then we will answer you concisely.

I don't know

I didn't see how you looked, and I didn't see how you used it.

... ... If you ask a concise question next time, then we will answer you concisely.

I don't know

I didn't see how you looked, and I didn't see how you used it.

$sql = "Select Djs_skupdate_class.id,djs_skupdate_class.grouptypeid,djs_skupdate_class.teachername,djs_skupdate_ Class.years,djs_skupdate_class.saydate,djs_skupdate_class.isupdate,djs_skgrouptype.groupname from Djs_skupdate_ Class,djs_skgrouptype where Djs_skupdate_class.grouptypeid = Djs_skgrouptype.id "; $sql =" SELECT * FROM Djs_skupdate_ Class ";//print_r ($sql); $dsql->execute (' Me ', $sql);            while ($arr = $dsql->getarray (' Me ')) {$row []=array (' id ' = ' = ' $arr ' grouptypeid ' = '],//' + $arr [' Grouptypeid '], ' GroupName ' + $arr [' groupname '], ' teachername ' + $arr [' teachername '], ' years ' = $arr [' Years '], ' saydate ' = > $arr [' saydate '], ' isupdate ' = $arr [' isupdate ']);            }

 
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Is this the SQL command you executed?
$sql = "SELECT * from Djs_skupdate_class";

Where did you get the multi-table?

Is this the SQL command you executed?
$sql = "SELECT * from Djs_skupdate_class";

Where did you get the multi-table? But I wrote that in SQL that they could find out two of the information on a table.

You are only $sql = "Select Djs_skupdate_class.id,djs_skupdate_class.grouptypeid,djs_skupdate_class.teachername,djs_ Skupdate_class.years,djs_skupdate_class.saydate,djs_skupdate_class.isupdate,djs_skgrouptype.groupname from Djs_ Skupdate_class,djs_skgrouptype where Djs_skupdate_class.grouptypeid = Djs_skgrouptype.id ";
This one?
But he was not under the
$sql = "SELECT * from Djs_skupdate_class";
Is it covered?

Two SQL.

$you = ' Zhang San ';
$you = ' John Doe ';
You are Li Four, not Zhang San.

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