从数据库中检索行并存储到会话数组[重复]

Question

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• PHP parse/syntax errors; and how to solve them? 17 answers

I am trying to retrieve rows from a database and store them to an array/ session array (I am a bit lost). The database I am currently retrieving from has 3 rows.

<?php
session_start();
$user_year = $_SESSION['year'];
$floor = $_SESSION['year_floor'];

include "config.php";
$query = "select * from timetable where '$user_year' = year;
$array = array();

while($row = $query->fetch_assoc()){
  $array[] = $row;

}

  echo '<script>console.log(\"$array\")</script>';
  /* close connection */
  // debug:
  print_r($array); // show all array data
  echo $array[0]['username']; // print the first rows username
  $mysqli->close();
?>

This is what I have pieced together so far, is this close? Any pointers in the right direction would be great thanks.

</div>

Answer 1

It's close but not quite. Check the examples at

http://php.net/manual/en/mysqli.query.php and

http://php.net/manual/en/mysqli-result.fetch-assoc.php and

http://php.net/manual/en/mysqli-result.fetch-all.php

Edits below.

I would also consider looking up Binding variables instead of injecting session variables into a query. A malicious user could potentially set your $user_year variable and edit the query. More here: http://php.net/manual/en/mysqli-stmt.bind-param.php

<?php
session_start();
$user_year = $_SESSION['year'];
$floor = $_SESSION['year_floor'];

include "config.php"; //I assume you create a $mysqli object here
$query = "select * from timetable where year = '$user_year'";
$results_array = array(); // I try to avoid using variable types as names so have renamed to results_array

$result = $mysqli->query($query); // mysqli runs the query (see man above)

//The result runs the fetch_assoc (see man above)
while($row = $result->fetch_assoc()){
     $result_array[] = $row;
}

// If you know you have a small result set you could replace the while() loop above with $result_array = $result->fetch_all()

// echo '<script>console.log(\"$array\")</script>'; (this wont work because the client side javascript can't understand the php variable. But you could place this line inside the loop above.)

// debug:
print_r($result_array); // show all array data
echo $result_array[0]['username']; // print the first rows username
$mysqli->close(); // you don't need to do this unless saving memory mid script, but it's good practice so I left it in
?>