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gorilla0

Interesting "bug"

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I noticed something quite strange & wondering if anyone has come across this:

I created a record set & wanted to make the table_name a variable. I declared the variable, but Dreamweaver wanted to parse it as just a string rather than a variable. I tested to see if the variable was set properly and it was. Here is an example of what I mean:

 

$table = \"colors\";

 

$sql = \"select * from $table\";

 

it tries to find a table name called \"$tables\" rather than \"colors\". This is VERY VERY basic stuff, I am really suprised at such a sloppy error. Anybody seen this?

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its not an error, single-quote the variable as the table name

 

[php:1:2fe72d1b6e]

$table = \"roles\";

mysql_query(\"SELECT * FROM \'$table\' \");

[/php:1:2fe72d1b6e]

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Thanks, if the solution works, it will save some redundant work, but I have to say:

Technically, it is a bug, since within regular PHP coding there is absolutely no need to enclose a table_name in single quotes. In fact, I know of an instance where it needs to be done, whether it me MS SQL, Oracle or PostGres. I am a six - seven-month newbie to Studio MX, but not to PHP development by a longshot, so I would definitely consider this a DreamWeaver bug. MacroMedia Gods, here me :-P

 

Additionally, I have not tried your fix, but am curious of the result.

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how is this a dreamweaver bug? lol

 

you write the code, php does it, mysql holds the data php uses, simple

 

dreamweaver is just a program we use to write the code. try the same code in notepad or any other editor, its the languages limitations.

 

a dreamweaver bug is if something in the GUI glitches, or if it doesn\'t display your code as it should is design view, or vice versa...the possibilities go on

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Let me be specific:

 

$table_name = \"my_table\";

$sql = \"select * from $table_name\";

$res = mysql_query($sql);

 

This WORKS all day long in averyday PHP.

 

in DreamWeaver\'s Recordset function, if you edit the code manually to reflect the same thing, it errors and says that $table_name does not exist.(We are not looking for $table_name, we are looking for the value of $table_name, that being \"my_table\") In other words, it views the $ as a regular character rather than understanding that it is a variable. Is that more clear? Sorry if I was not as specific in my first example.

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