php - 如何在 php 中将 tinyint 值更新为 1 而不是变为 NULL

Question

我正在尝试将数据库布尔值（tinyint）更新为 1，但相反，它正在更新为 NULL，并且所有其他值都在正确更改。

当我离开它时得到的错误如下：

绑定失败：0 执行失败：2031 未为准备好的语句中的参数提供数据

如果我改变 $dealt_Out = 1; 对于一个变量，在准备好的语句之前，所有值都被正确更改，除了布尔值，它最终为 NULL。

<?php
if(!($stmt = $mysqli->prepare("UPDATE card SET player_Holding = ?, dealt_Out = 1 WHERE id_Card = ?"))){ 
echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
}
if(!($stmt->bind_param("iii",$_POST['player_Holding'],$_POST['dealt_Out'],$_POST['id_Card']))){

echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
echo "Execute failed: "  . $stmt->errno . " " . $stmt->error;
}     else {
echo "Edit " . $stmt->affected_rows . " row to Card Holder. <br/><br/><strong>Returning to 'Make a Move'</strong>";
}

?>

Answer 1

检查好。应该是delt_out = ? not delt_out = 1 如果不是，则取出$_POST['delt_out']，在 bind_param 并删除一个i

我已经修改了你的代码。希望它能解决你的问题

<?php
$dealtout=1;
if(!($stmt = $mysqli->prepare("UPDATE card SET player_Holding = ?, dealt_Out = ? WHERE id_Card = ?"))){ 
echo "Prepare failed: "  . $stmt->errno . " " . $stmt->error;
}
if(!($stmt->bind_param("iii",$_POST['player_Holding'],$dealtout,$_POST['id_Card']))){

echo "Bind failed: "  . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
echo "Execute failed: "  . $stmt->errno . " " . $stmt->error;
}     else {
echo "Edit " . $stmt->affected_rows . " row to Card Holder. <br/><br/><strong>Returning to 'Make a Move'</strong>";
}

?>