Question

尝试在Laravel中使用mysql_fetch_array（）时出错。

while ($row = mysql_fetch_array($query))

使用什么而不是mysql_fetch_array（）？

我遇到错误，

mysql_fetch_array（）期望参数1是资源，给定数组

完整部分，

//prepare the tag cloud array for display
    $terms = array(); // create empty array
    $maximum = 0; // $maximum is the highest counter for a search term

    $query = DB::select('SELECT term, counter FROM search ORDER BY counter DESC LIMIT 30');

    while ($row = mysql_fetch_array($query))
    {
        $term = $row['term'];
        $counter = $row['counter'];

        // update $maximum if this term is more popular than the previous terms
        if ($counter > $maximum) $maximum = $counter;

        $terms[] = array('term' => $term, 'counter' => $counter);

    }

Answer 1

DB语句和选择返回结果，而不是查询对象，因此您只需要遍历结果。

$results = DB::select('SELECT term, counter FROM search ORDER BY counter DESC LIMIT 30');

foreach($results as $row)
{
    $term = $row->term;
    $counter = $row->counter;

    // update $maximum if this term is more popular than the previous terms
    if ($counter > $maximum) $maximum = $counter;

    $terms[] = array('term' => $term, 'counter' => $counter);

}

有关通过数据库类运行原始查询的更多信息，请参见here。

Answer 2

由于使用了关联数组从数据库中提取数据，因此您还可以在mysql_fetch_array（）中指定一个标志，即MYSQL_ASSOC。

<?php
$con = mysql_connect("localhost", "peter", "abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
$db_selected = mysql_select_db("test_db",$con);
$sql = "SELECT * from Person WHERE Lastname='Refsnes'";
$result = mysql_query($sql,$con);
print_r(mysql_fetch_array($result,MYSQL_ASSOC));
mysql_close($con);
?>

有关更多详细信息，请尝试：http://w3schools.sinsixx.com/php/func_mysql_fetch_array.asp.htm

Laravel不能使用mysql_fetch_array（）

2 个答案: