Question

编辑：解决，见下文但仍然不知道我做了什么让它工作：）

好吧，我现在被困住了。我有表users：

ID int PRIMARY AUTO_INCREMENT
EMAIL varchar(60)
NICK varchar(60)
//...

如果我这样做：

<?php
$email = $_POST["mai"];
$nickname = $_POST["nck"];
$mysqli = new mysqli($db_host, $db_username, $db_password, $database);
$prepared_statement = "INSERT INTO users VALUES(?,?,?)";
if ($stmt = $mysqli->prepare($prepared_statement)) {
 $id = "";
 $stmt->bind_param("iss",$id,$email,$nickname); 
 $stmt->execute();
}
?>

每次都有效。但如果我选择同样的话：

<?php
 $previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
 $check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=?;");
 $check->bind_param("s",$email);
 $check->execute();
 $check->bind_result($maybe_id,$maybe_got_something);

  while ($check->fetch()) { //here was typo, but fixed now 
    if ($maybe_got_something==$nickname){
     echo "Hooray!";
    }
  } 

?>

我从未见过“万岁！”

但是，如果我改变它：

$previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
$prepared_statement = "SELECT ID,NICK FROM users WHERE EMAIL=";
$prepared_statement .=$email;
$result = $previous_entries->query($prepared_statement);
while ($row = $result->fetch_array()){
  if ($row["NICK"]==$nickname){
     echo "Hooray!";
    }
}

然后一切都好。

我在准备好的声明中犯了一些可怕的错误。但是我真的找不到它......我在这里做错了什么？

编辑：更新了脚本以纠正错误的拼写错误，并添加了这两行：

echo "maybeid: ".. $maybe_id;
echo "maybenick:". $maybe_got_something;

Page echos this：

  maybeid: maybenick:

编辑：工作代码 在尝试调试时，我得到了这个：

$previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
$check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=?;");
$check->bind_param("s",$email);
$check->execute();
$check->bind_result($maybeid,$maybenick);
echo "maybeid: ".$maybeid;
echo "maybenick:". $maybenick;
// got rid of the if statement
  while ($check->fetch()) {
    echo "maybeid: ".$maybeid;
    echo "maybenick:". $maybenick;
    if ($maybenick==$nickname){

     echo "Hooray!";
    }
  }

但是......为什么这么令人担忧？：）

Answer 1

准备好的查询为$check，但您在提取时使用的是$stmt。

Answer 2

试试这个（使用从prepare返回的语句对象）：

<?php
 $previous_entries = new mysqli($db_host, $db_username, $db_password, $database);
 $check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=?;");
 $check->bind_param("s",$email);
 $check->execute();
 $check->bind_result($maybe_id,$maybe_got_something);
 if($check->num_rows > 0){
  while ($check->fetch()) {
    if ($maybe_got_something==$nickname){
     echo "Hooray!";
    }
  } 
}
?>

Answer 3

尝试使用此$stmt->fetch()

更改$check->fetch()

您可能需要将error_reporting级别设置为

error_reporting = E_ALL＆amp; 〜E_NOTICE

在你的php.ini for development环境中避免这种错误。

此致

Answer 4

您必须更改以下代码：

...
$check = $previous_entries->prepare("SELECT ID,NICK FROM users WHERE EMAIL=:email;");
$check->bind_param(":email",$email);
...

Prepared语句适用于INSERT但不适用于SELECT

4 个答案: