如何在PHP中使用数组进行MYSQL查询?
如果先前查询的结果在数组中,如何使用它们查询数据库?以下是我到目前为止的情况如何在PHP中使用数组进行MYSQL查询?,php,mysql,arrays,Php,Mysql,Arrays,如果先前查询的结果在数组中,如何使用它们查询数据库?以下是我到目前为止的情况 $query = "SELECT * from searchtestdb where engname in ( SELECT synonyms.synonym FROM words LEFT JOIN synonyms ON synonyms.word_id = words.word_id WHERE word LIKE '%$searchBox%') "; while($result = mysql_fetch
$query = "SELECT * from searchtestdb where engname in ( SELECT synonyms.synonym FROM words LEFT JOIN synonyms ON synonyms.word_id = words.word_id WHERE word LIKE '%$searchBox%') ";
while($result = mysql_fetch_array($query))
{
echo $result['engname'];
echo "<br> ";
echo "<br> ";
}
但这显然不起作用。有什么想法吗?。它们不再得到维护。看到了吗?相反,学习,并使用or-将帮助您决定哪一个。如果选择PDO,。为什么要再次查询相同的表以获得相同的行?您是打算使用$result['engname']查找与engGame相似的行(因此类似),还是希望找到相同的行?答案很好,但我认为有必要对concating字符串进行更多解释:有关更多信息,请阅读以下内容:
while($result = mysql_fetch_array($query))
{
$query2 = "SELECT * from searchtestdb where engname LIKE '%".$result['engname']."%';";
$result2 = mysql_query($query2);
while($row = mysql_fetch_array($result2))
{
echo $row["somthing"];
}
}
while($result = mysql_fetch_array($query))
{
$query2 = "SELECT * from searchtestdb where engname LIKE '%".$result['engname']."%';";
$result2 = mysql_query($query2);
while($row = mysql_fetch_array($result2))
{
echo $row["somthing"];
}
}