Question

我想从我的数据库中读取所有数据，当我从控制台运行此查询时，我得到了很多结果，但由于某种原因，php只读了一个。

$query = "
   SELECT b.raw as 'address', a.raw as 'name', c.TouchTime as 'time'
   FROM touchName a, touchHome b, trackTouch c
   WHERE a.raw like \"%{$name}%\" 
      AND c.AgentID = 1
      AND a.relations = b.relations
      AND b.relations = c.relations
      AND a.relations = c.relations
   ORDER BY time desc
";

//So we can double check in the console
echo $query . "<br><br>";

$result = mysqli_query($mc, $query);

$array = mysqli_fetch_assoc($result);

//says there is only one row
$total = count( mysqli_num_rows($result) );
echo $total."<br>";

我已经尝试了很多方法来破坏结果中的数据，我已经通过多种方式修改了查询，包括分组，计数等等，试图将其分解。

加入时也相当新，如果这很难看，那是因为这是第一次没有吐出1200万个结果的黑客。

Answer 1

试试这个：

$array = array();
while ($row = mysql_fetch_assoc($result)) {
   $array[] = $row;
}

Answer 2

函数mysql_fetch_assoc默认返回单个记录。你需要迭代循环，因为@tyralcori说..

Answer 3

$query = "
   SELECT b.raw as 'address', a.raw as 'name', c.TouchTime as 'time'
   FROM touchName a, touchHome b, trackTouch c
   WHERE a.raw like \"%{$name}%\" 
      AND c.AgentID = 1
      AND a.relations = b.relations
      AND b.relations = c.relations
      AND a.relations = c.relations
   ORDER BY time desc
";

echo $query;
$handler= mysqli_query($mc, $query);
$result = array();  
while($rec = mysqli_fetch_assoc($handler)){
    $result[]['address'] = $rec['address'];
    $result[]['name'] = $rec['name'];
    $result[]['time'] = $rec['time'];
}
print_r($result);

PHP只从mysql查询返回一个结果 - mysql控制台返回数十个

3 个答案: