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How can i change my form below, so that the chosen option populates the $user_id variable.

I would like to use he variable in an sql query.

 

eg. user selects option value 5, then $user_id = 5,

 

name: 'S Wedge',
data: [<?php
$sql= "SELECT month_table.month , IFNULL(AVG( yard ), 0) AS avscore FROM month_table LEFT JOIN `snag` ON month_table.month =
 MONTH( STR_TO_DATE( sdate, '%d/%m/%y' ) ) AND club = 'S wedge' AND user_id = $user_id AND YEAR = $year GROUP BY `month`";
$result=mysql_query($sql);
while($row = mysql_fetch_array($result) ){ echo $row['avscore'].',';}
?>]
}]
});
});

});
</script>
</head>
<body>
<script src="../../js/highcharts.js"></script>
<script src="../../js/modules/exporting.js"></script>


<form action="action=<?php echo htmlentities($_SERVER['PHP_SELF']); ?> method="post">"
<label for='user_id[]'>Select the user you would like to display:</label><br>
<select multiple="multiple" name="user_id[]">

<option>Select User</option>
<option value="1">1</option>
 <option value="3">3</option>
 <option value="5">5</option>
<option value="7">7</option>

</select>
<input type="submit" value="Submit">
</form>

<?php echo $user_id ?>

<div id="container" style="min-width: 400px; height: 400px; margin: 0 auto">
</div>

</body>
</html>

Link to comment
https://forums.phpfreaks.com/topic/271775-using-a-form-to-select-variables/
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Yes, all you need to do is to read up on the $_POST superglobal. As well as intval (), to ensure that you are indeed getting an integer (and not an SQL injection attack).

 

PS: I recommend using this snippet for protecting yourself against XSS attacks via $_SERVER['PHP_SELF'].

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