我有几个查询字符串,我想使用" mysqli_multi_query"一次执行。这很有效。
当我再次插入查询以使用" mysqli_query"检查连接表中的每个项目时它不会从PHP返回任何结果或任何错误。当我在phpmyadmin中手动运行查询字符串时,一切正常。
这是我的代码:
<?php
$connect = mysqli_connect('localhost','root','','database');
$strquery = "";
$strquery .= "1st Query";
$strquyer .= "2nd Query";
if($multi = mysqli_multi_query($connect,$strquery)){ // function mysqli_multi_query is working
// From here it doesn't give any response
$qryarray = mysqli_query($connect,
"SELECT purchase_detail_$_SESSION[period].item_code,
purchase_detail_$_SESSION[period].location_code
FROM purchase_detail_$_SESSION[period]
WHERE purchase_detail_$_SESSION[period].purchase_num = '$_POST[purchase_num]'
UNION
SELECT purchase_detail_temp.item_code,
purchase_detail_temp.location_code
FROM purchase_detail_temp
WHERE purchase_detail_temp.purchase_num = '$_POST[purchase_num]' AND purchase_detail_temp.username = '$_SESSION[username]'");
while($array = mysqli_fetch_array($qryarray)){
"Some code to process several item code in table purchase_detail_$_SESSION[period]"
}
}
我的代码有什么问题吗?
答案 0 :(得分:8)
感谢@Fred -ii-提供您的信息,以便我可以解决问题...
我刚刚在php manual =
中找到答案WATCH OUT: if you mix $mysqli->multi_query and $mysqli->query, the latter(s) won't be executed!
BAD CODE:
$mysqli->multi_query(" Many SQL queries ; "); // OK
$mysqli->query(" SQL statement #1 ; ") // not executed!
$mysqli->query(" SQL statement #2 ; ") // not executed!
$mysqli->query(" SQL statement #3 ; ") // not executed!
$mysqli->query(" SQL statement #4 ; ") // not executed!
The only way to do this correctly is:
WORKING CODE:
$mysqli->multi_query(" Many SQL queries ; "); // OK
while ($mysqli->next_result()) {;} // flush multi_queries
$mysqli->query(" SQL statement #1 ; ") // now executed!
$mysqli->query(" SQL statement #2 ; ") // now executed!
$mysqli->query(" SQL statement #3 ; ") // now executed!
$mysqli->query(" SQL statement #4 ; ") // now executed!
所以我只需在mysqli_multi_query()
之后插入此代码:
而(mysqli_next_result($连接)){;}
希望它有用..谢谢..
答案 1 :(得分:5)
除了爱迪生的回答:
而(mysqli_next_result($连接)){;}
我会推荐这段代码。这避免了例外。
while(mysqli_more_results($con))
{
mysqli_next_result($con);
}