我在我的网络服务器上使用 SQLite，直到现在都没有问题。
$sql = "SELECT * from TeammateCurrent;";
$ret = $db->query($sql);
if($ret != false) {
while($row = $ret->fetchArray(SQLITE3_ASSOC) ){
$uniqueid = $row['uniqueID'];
$name = $row['Name'];
$lineID = $row['LineID'];
$job = $row['Job'];
$sunday = $row['Sunday'];
$monday = $row['Monday'];
$tuesday = $row['Tuesday'];
$wednesday = $row['Wednesday'];
$thursday = $row['Thursday'];
$friday = $row['Friday'];
$saturday = $row['Saturday'];
//echo $name."<br>".$lineID."<br>".$job."<br>".$sunday."<br>".$monday."<br>".$tuesday."<br>".$wednesday."<br>".$thursday."<br>".$friday."<br>".$saturday."<br><br>";
$sql = "INSERT INTO TeammateHistory (uniqueID, Name,LineID,Job,Sunday,Monday,Tuesday,Wednesday,Thursday,Friday,Saturday)
VALUES (NULL, '$name','$lineID','$job','$sunday','$monday','$tuesday','$wednesday','$thursday','$friday','$saturday');";
$ret = $db->exec($sql);
if(!$ret){
echo $db->lastErrorMsg();
} else {
}
$sql2 = "UPDATE TeammateCurrent set Sunday='$day0', Monday='$day1', Tuesday='$day2', Wednesday='$day3', Thursday='$day4', Friday='$day5', Saturday='$day6' where uniqueID='$uniqueid';";
$ret = $db->exec($sql2);
if(!$ret){
echo $db->lastErrorMsg();
} else {
}
}
} else {
echo "Query not successful";
}
当上面的代码执行时，我得到错误
Fatal error: Call to a member function fetchArray() on a non-object in /home/xxxx/public_html/xxxx.com/trg/rc2/db/startNewWeek.php on line 39我在其他页面上运行过这样的代码，它们工作正常，我在这里完全不知所措。
我的 table :
最佳答案
您正在重用变量 $ret . SELECT 语句和 UPDATE 应该有不同的返回变量。在第二个循环中，您尝试执行 fetchArray() UPDATE 语句的返回。
关于php - fetchArray() 在 SQLite 中不起作用，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37923063/