关于为什么我的数据没有更新的任何想法?他们写我准备好的陈述的方式是他们的根本错误吗?
表格:
while($log_dates = mysqli_fetch_assoc($q_log_dates_result)) {
echo "<tr>";
echo "<input type='hidden' name='data[][log_dates_ID]' value='" . $log_dates['log_dates_ID'] . "'/>";
echo "<td><input type='text' name='data[][week_date]' value='" . $log_dates['week_date'] . "' /></td>";
echo "<td><input type='text' name='data[][crew_chief]' value='" . $log_dates['crew_chief'] . "' readonly /></td>";
echo "<td><input type='text' name='data[][monday_crew]' value='". $log_dates["monday_crew"] ."'/></td>";
echo "</tr>";
} // end while loop
PHP:
if (isset($_POST['submit'])) {
$stmt = $connection->stmt_init();
if($stmt->prepare("UPDATE log_dates SET (week_date, crew_chief, monday_crew) VALUES (?, ?, ?) WHERE log_dates_ID = ?")) {
// Bind your variables to replace the ?s
$stmt->bind_param('sssi', $week_date, $crew_chief, $monday_crew, $log_dates_ID);
$returnedData = $_POST['data'];
for($i=0;$i<count($returnedData);$i+=4){
$log_dates_ID = $returnedData[$i]['log_dates_ID'];
$week_date = $returnedData[$i+1]['week_date'];
$crew_chief = $returnedData[$i+2]['crew_chief'];
$monday_crew = $returnedData[$i+3]['monday_crew'];
$stmt->execute();
}
// Close statement object
$stmt->close();
}
}
答案 0 :(得分:2)
您的UPDATE语法不正确。它应该是:
UPDATE log_dates
SET week_date = ?, crew_chief = ?, monday_crew = ?
WHERE log_dates_ID = ?
您尝试在INSERT
语句中使用UPDATE
语法。它们根本不相似。
答案 1 :(得分:1)
我会这样做......
在函数内......
function updatedata($week_date, $crew_chief, $monday_crew, $log_dates_ID){
$stmt = $connection->stmt_init();
$stmt->prepare("UPDATE log_dates SET (week_date, crew_chief, monday_crew) VALUES (?, ?, ?) WHERE log_dates_ID = ?"))
$stmt->bind_param($week_date, $crew_chief, $monday_crew, $log_dates_ID);
$stmt->execute();
}
然后在你的循环中......
for($i=0;$i<count($returnedData);$i+=4){
$log_dates_ID = $returnedData['log_dates_ID'][$i];
$week_date = $returnedData['week_date'][$i];
$crew_chief = $returnedData['crew_chief'][$i];
$monday_crew = $returnedData['monday_crew'][$i];
updatedata($week_date, $crew_chief,$monday_crew, $log_dates_ID);
}