再次问好 [英] Hello again

问题描述

再次帮助


这个工作


$ q =" SELECT * FROM groupnames WHERE EXISTS（select * FROM tenets WHERE

groupnames.groupnum = tenets.groupnum AND tenets.level< =''1''）" ;;


这不是


$ q =" SELECT * FROM groupnames WHERE EXISTS（select * FROM tenets WHERE

groupnames.groupnum = tenets.groupnum AND tenets.level< =

$ session-> userleve）" ;;


我似乎错过了关于连接的内容


Len Bell <登记/>

解决方案

q =" SELECT * FROM组名WHERE EXISTS（SELECT * FROM原则WHERE

groupnames.groupnum =原则。 groupnum和tenets.level< =''1''）" ;;


这不是

q =" SELECT * FROM groupnames WHERE EXISTS（select * FROM tenets WHERE

groupnames.groupnum = tenets.groupn嗯和tenets.level< =

session-> userleve）" ;;


我好像错过了什么关于串联


Len Bell

Help again

This works

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= ''1'')";

This doesn''t

$q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=
$session->userleve)";

I seem to have missed something about concatenation

Len Bell

解决方案

q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <= ''1'')";

This doesn''t

q = "SELECT * FROM groupnames WHERE EXISTS (select * FROM tenets WHERE
groupnames.groupnum = tenets.groupnum AND tenets.level <=

session->userleve)";

I seem to have missed something about concatenation

Len Bell

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