問題描述

我的數據庫中有一個存儲過程，它返回表中的所有記錄:

I have a stored procedure in my db that returns all records in a table:

CREATE PROCEDURE showAll()
BEGIN
    SELECT * FROM myTable;
END

SP 工作正常.但是，如果我在 php 腳本中調用它然后我再次嘗試查詢數據庫，它總是失敗:

The SP works just as expected. But, if I call it in a php script and then I try to query the database again, it always fail:

// $mysqli is a db connection

// first query:

if (!$t = $mysqli->query("call showAll()"))
    die('Error in the 1st query');

while ($r = $t->fetch_row()) {
    echo $r[0] . "<br>"; // this is ok
}

$t->free(); // EDIT (this doesn't help anyway)

// second query (does the same thing):

if (!$t = $mysqli->query("SELECT * from myTable"))
    die('Error in the 2nd query'); // I always get this error

while ($r = $t->fetch_row()) {
    echo $r[0] . "<br>";
}

值得注意的是，如果我交換兩個查詢(即我在最后調用存儲過程)，它可以正常工作而不會出現任何錯誤.在第二個查詢之前關閉()結果沒有幫助.一些提示?

Notable, if I swap the two queries (i.e. I call the stored procedure at the end) it works without any error. To close() the result before the second query doesn't help. Some hints?

mysqli::error() 是:?命令不同步；您現在無法運行此命令?.

mysqli::error() is: ?Commands out of sync; you can't run this command now?.

推薦答案

mysqli.query 的 php.net/manual 條目的評論已更新，現在包含對此的答案.總而言之，在 $t->close() 之后調用 $mysqli->next_result().向 petrus.jvr 致敬！

The comments on the php.net/manual entry for mysqli.query have been updated, and now include an answer for this. To summarize, call $mysqli->next_result() after $t->close(). Kudos to petrus.jvr!

鏈接:http://www.php.net/manual/en/mysqli.query.php#102904

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