Question

我正在传递JSON输入，然后解码值，然后将值传递给SQL查询，以根据我提到的条件从表中获取数据。但对我来说，当我在网页中打开PHP脚本时，它以JSON格式显示表的所有值。此外，我没有从我的输入中获取DB的值。之前这个脚本正在运行。这个脚本有什么问题。

$jsonString ='{"imdblist":{},"rottenlist":{},"generelist":{"genere1":"Documentary"},"yearlist":{}}';
$obj = json_decode($jsonString,true);

$imdb=array();
$rotten=array();
$genere=array();
$year=array();

foreach ($obj['imdblist'] as $key => $value) 
    {
       echo "<br>------" . $key . " => " . $value;
        $imdb[$key] = $value;

    }
foreach ($obj['rottenlist'] as $key => $value) 
    {
       // echo "<br>------" . $key . " => " . $value;
        $rotten[$key] = $value;

    }   
 foreach ($obj['yearlist'] as $key => $value) 
    {
        //echo "<br>------" . $key . " => " . $value;
        $year[$key] = $value;

    }

$val1=implode("','", $imdb);
$val1 = "'" . $val1 . "'";
$val2=implode("','", $rotten);
$val2 = "'" . $val2 . "'";
$val4=implode("','", $year);
$val4 = "'" . $val4 . "'";

array_walk($obj['generelist'],function(&$item1, $key){$item1="$item1=1";});
$stringgen = implode(' OR ', $obj['generelist']);

if(empty($stringgen)){
    $myquery = "SELECT * FROM Moviess WHERE `imdb_rat` IN ($val1) OR `Rotten_rate` IN ($val2) OR `Year` IN ($val4)";
      } 
 else {
   $myquery = "SELECT * FROM Moviess WHERE `imdb_rat` IN ($val1) OR `Rotten_rate` IN ($val2)  OR ".mysql_real_escape_string($stringgen)." OR `Year` IN ($val4)";
  }

$query = mysql_query($myquery);

if ( ! $query ) {
    echo mysql_error();
    die;
}    

for ($x = 0; $x < mysql_num_rows($query); $x++) {
    $data[] = mysql_fetch_assoc($query);
}

echo json_encode($data);

在上面的查询中，我只需要获取Documentary的内容，但是当我运行查询时不会获取数据。如果我在网页中打开它，我还能够以JSON格式查看表中的所有值。这段代码有什么问题？

php sql查询从db表android获取所有值，即使有传递查询的条件

0 个答案: