Question

我试图使用预准备语句在数据库中输入数据。毫无准备的陈述有效，但这份准备好的陈述没有。我无法找出原因。准备版本：

birthtown = StringField("Birthtown", filters = [lambda x: x or None])

未准备好的版本：

$stmt = $mysqli->prepare("INSERT INTO videos (file_name, upload_by, date, path) 
VALUES (?, ?, ?, ?)");
$stmt->bind_param('ssss', $newstring, $id, $date->format('Y-m-d'), $location);
$stmt->execute();

Answer 1

将$stmt-execute();替换为$stmt->execute();

另外，请勿在查询中使用date和path。使用其他名称对其进行重命名，例如date1和path1。

更新您的查询，如下所示，肯定会有效（离线测试）：

<?php
$mysqli = new mysqli('localhost', 'root', '', 'test2'); 

if ($mysqli->errno) {
printf("Connect failed: %s\n", $mysqli->error);
exit();
}

$stmt = $mysqli->prepare("INSERT INTO videos (file_name, upload_by, date1, path1) VALUES (?, ?, ?, ?)");
$stmt->bind_param('ssss', $file_name, $upload_by, $date1, $path1);
$date1 = date("Y-m-d");
$file_name = "test.jpg";
$upload_by = "amit";
$path1 = "test";
if ($result = $stmt->execute()){
echo "success";
$stmt->free_result();
} else {
echo "error";
}
$stmt->close();
?>

Answer 2

如果您只使用？，则将参数绑定两次？，不要再次绑定参数直接执行。

 //Prepare your query first
 $stmt = $mysqli->prepare("INSERT INTO videos (file_name, upload_by, date, path) 
    VALUES (?, ?, ?, ?)");

//Just pass your argument and execute directly without binding the parameter (The parameter is binded already)
$stmt->execute('ssss', $newstring, $id, $date->format('Y-m-d'), $location);

Mysqli绑定参数不起作用

2 个答案: