我准备的声明更新存在问题
$query = "UPDATE kids_entry SET entries=? WHERE parentsemail=?";
$stmt1 = mysqli_prepare($connection, $query);
mysqli_stmt_bind_param($stmt1, 'is',$entries,$parentsemail);
mysqli_execute($stmt1);
if(mysqli_stmt_affected_rows($stmt1) != 1)
die("issueasdass");
mysqli_stmt_close($stmt1);
上面的代码在另一个页面上确实对我有用，但我似乎无法让它在这里工作。在另一页上，要更新的设置是硬编码的并且不是动态的，因此这可能就是它对我不起作用的原因。有人可以告诉我我做错了什么吗？谢谢
$query = "UPDATE kids_entry SET entries=? WHERE parentsemail=?";
$stmt1 = mysqli_prepare($connection, $query);
mysqli_stmt_bind_param($stmt1, 'is',$entries,$parentsemail);
mysqli_execute($stmt1);
if(mysqli_stmt_affected_rows($stmt1) != 1)
die("issueasdass");
mysqli_stmt_close($stmt1);
The above code does work for me on another page but i cannot seem to get it to work here. On the other page the set to update is a hardcoded in and is not dynamic so that might be why it isn't working for me. Can someone tell me what I am doing wrong? Thanks
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条目的值是否可能与数据库中已有的值相同？我认为如果没有任何变化，
*_affected_rows不会返回 1。是否可能有超过 1 个“parentsemail”条目？那么也许 mysqli_stmt_affected_rows 会返回类似 2 的内容？也许让我们知道 mysql_stmt_affected_rows 正在返回什么，而不是检查 1。例如，来自文档：Is it possible that entries is the same value as what's already in the database? I don't think
*_affected_rowsreturns 1 if nothing changes. Is it possible that there's more than 1 entry for 'parentsemail'? So perhapsmysqli_stmt_affected_rowsis returning something like 2? Instead of checking 1, maybe let us know whatmysql_stmt_affected_rowsis returning. For example, from the docs:这个问题似乎已经自行解决。没有任何改变，但现在可以了！
The issue appears to have resolved itself. Nothing has changed but now it works!