由于一些奇怪的原因,查询的准备失败,我检查了一万次,参数是正确的以及表。这是我的代码:
zipOut.write(fileNameAndPath, actualFile)
}
我尝试在else对象中进行打印,但我只得到了
这个词执行()失败:
foreach($decoded as $caption)
{
if($stmt = $mysqli->prepare("SELECT id, leagueCode FROM league"))
{
$stmt->bind_result($id, $leagueCode);
$stmt->execute();
var_dump($stmt);
while($stmt->fetch())
{
if($leagueCode == $caption['league'])
{
if ($stmt = $mysqli->prepare("INSERT INTO soccerseason (id, caption, league, years, numberOfTeams, numberOfGames, lastUpdated, self, teams, fixtures, leagueTable) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"))
{
$id_fk = $id;
$caption = $caption['caption'];
$league = $caption['league'];
$years = $caption['year'];
$not = $caption['numberOfTeams'];
$nog = $caption['numberOfGames'];
$lu = $caption['lastUpdated'];
$self = $caption['_links']['self']['href'];
$teams = $caption['_links']['teams']['href'];
$fix = $caption['_links']['fixtures']['href'];
$lt = $caption['_links']['leagueTable']['href'];
if ($stmt->bind_param("isssiisssss", $id_fk, $caption, $league, $years, $not, $nog, $lu, $self, $teams, $fix, $lt))
{
if ($stmt->execute())
{
echo "Done!";
}
else
{
die($mysqli->error);
}
}
else
{
die($mysqli->error);
}
}
else
{
die('execute() failed: ' . $mysqli->error);
}
}
}
}
似乎else
{
die('execute() failed: ' . $mysqli->error);
}
为空,因为它没有打印,为什么会发生这种情况?我做错了什么?
答案 0 :(得分:0)
我终于找到了问题,并且我意识到如果由于某种原因使用相同的连接,对象的参数将丢失,因此在执行相同的参数后必须使用$stmt-> store_result();
。
我解决了这个问题,希望我的回答能有所帮助。