mysqli预处理语句其中var = var dynamic

Question

If I have a prepared statement as follows:

$stmt = $mysqli->prepare( "SELECT fielda, fieldb, fieldc, from tablea where $option = ?" )

Is it possible to prepare the $option variable as well?

Note: the $option variable comes from a drop down list as follows

<select name="option">
  <option value="blah1">blah1</option>
  <option value="blah2">blah2</option>
  <option value="blah3">blah3</option>
  <option value="blah4">blah4</option>
</select>

and the other field comes from a simple input text box. This field will fill up the ? in the prepared statement.

Answer 1

you can use method "bindParam"

$stmt = $dbh->prepare("INSERT INTO REGISTRY (name, value) VALUES (?, ?)");  
$stmt->bindParam(1, $name); $stmt->bindParam(2, $value);