Brother, after I pass the program written in php to the client, there are related problems that cannot be browsed, and the catalog cannot be browsed.

After I passed the program written in php to the customer, there was a problem that I could not browse, and the directory could not be browsedAfter I passed the program written in php to the customer, there was a problem that I could not browse, and the directory could not be browsed, but the customer Has been set to 777 permission, still can not browse, please solve it. But php seems to be able to execute, but neither pictures nor css can be read——Solution—————- —-Looks like apache configuration problem The file does not give access——Solution ——————–It should be a configuration problem of nginx——Solution ——————–nginx needs to set up two directories, one for php scripts and another for html and css

Brother, I want to create a table in sql statement in php according to the parameters entered by the user

I want to create a table in the sql statement in php according to the parameters entered by the userFor example, I use $_GET[‘tablename’] to get the user’s input value, and I want to use the sql statement in the php file to create a name called The form of $_GET[‘tablename’], how to express the tablename herecreate table [tablename]? This sql statement is correctI use create table {. $_GET[‘tablename’ ]. } Can’t be realized? ——Solution——————- $sql = <<<SQLcreate table {$_GET[‘tablename’]}…SQL; ——Solution ——————–= <<<SQL is here documentheredoc syntax structure: <<<. An identifier is provided after the operator, followed by a newline. Next comes the string itself, ending with the previously defined identifier. <?php $cOntent= <<<FDIPZONE blog FDIPZONE; echo $content; ?> Reference: http://blog.csdn.net/fdipzone/article/details/24937669——Solution ———– ———That’s it. $sql =<<<SQLCREATE TABLE `{$_GET[‘tablename’]}` ( `id` int(10 ) unsigned NOT NULL auto_increment, `name` varchar(100) NOT NULL, `age` tinyint(4) unsigned NOT NULL, `addtime` datetime NOT NULL, PRIMARY KEY (`id`))SQL; mysql_query($sql) or die(mysql_error());

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