NLCJ Posted June 26, 2010 Share Posted June 26, 2010 Hello, I'm creating a new site, and to do that I will have to do it like below. Basically I have to display something that I got from SQL, but the row I need is a variable. How do I do this? I've tried all possibilities I could think of... $name = "NameOfSomething"; $sql = //MySQL query echo $sql[$name]; I hope you guys understand my problem. Thanks, Link to comment https://forums.phpfreaks.com/topic/205956-variable-in-mysql-command/ Share on other sites More sharing options...
Alex Posted June 27, 2010 Share Posted June 27, 2010 You need to post relevant code. Normally you'd do something like this: $sql = "..."; $result = mysql_query($sql); $row = mysql_fetch_assoc($result); echo $row[$name]; Link to comment https://forums.phpfreaks.com/topic/205956-variable-in-mysql-command/#findComment-1077800 Share on other sites More sharing options...
NLCJ Posted June 27, 2010 Author Share Posted June 27, 2010 Okay. Here is a part of the code, which I think is necessary: <?php $id = $_GET['id']; $table1 = mysql_fetch_array(mysql_query("SELECT * FROM table1 WHERE id='".mysql_real_escape_string($id)."'")); $table2 = mysql_fetch_array(mysql_query("SELECT * FROM table2 WHERE date='".mysql_real_escape_string($table1['date'])."'")); Lets say that there is a row with the same name as the result of table 1. echo $table2[$table1['name']]; ?> However this doesn't work... Thanks for your reply. Link to comment https://forums.phpfreaks.com/topic/205956-variable-in-mysql-command/#findComment-1077802 Share on other sites More sharing options...
NLCJ Posted June 27, 2010 Author Share Posted June 27, 2010 Somehow it's working now... I guess I was just to tired. Thanks for all the help though! Link to comment https://forums.phpfreaks.com/topic/205956-variable-in-mysql-command/#findComment-1077816 Share on other sites More sharing options...
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