使用准備好的語句在表中找不到名稱

Question

我正在嘗試從$_GET[]變量中搜索名稱，我正在使用准備好的語句，但由於某種原因，我找不到名稱，我測試了試圖找到 ID，並且它有效，但是對於字符串，它似乎不起作用。 我的數據庫中有我試圖搜索為VARCHAR的列。

]

Php 連接：

<?php
  $servername = "localhost";
  $username = "root";
  $password = "root";
  $dbname = "loginsistema";

  $sensor1 = $_GET["s1"]; //Trocar isto por POST
  $sensor2 = $_GET["s2"]; //Trocar isto por POST
  $sensor3 = $_GET["s3"]; //Trocar isto por POST 
  $chave = $_GET['chave']; //Trocar isto por POST 

  $conn = mysqli_connect($servername, $username, $password, $dbname);

  if ($conn->connect_error){
    die ("Connection failed". $conn->connect_error);
  }

  //Ver se a chave existe nos aparelhos, e para inserir os dados na tabela certa
  $sql = "SELECT * FROM aparelhos WHERE numdserie=?;";
  $stmt = mysqli_stmt_init($conn);
  if (!mysqli_stmt_prepare($stmt, $sql))
  {
    echo '<script> window.alert("Opa! MySQLI Error Cx!"); </script>";';
    exit();
  }
  else
  {
    mysqli_stmt_bind_param($stmt, "s", $chave);
    mysqli_stmt_execute($stmt);
    $result = mysqli_stmt_get_result($stmt);

    if (mysqli_num_rows($result) > 0)
    {
      $dbname = "infosensor";
      $conn = mysqli_connect($servername, $username, $password, $dbname);

      $sql = "INSERT INTO `1` (chave,sensor1,sensor2,sensor3) VALUES ($chave, $sensor1,$sensor2,$sensor3);";
    }

  }

  if($conn->query($sql) == TRUE){
    echo "New record created succesfully.</br>";
  }else{
    echo "Error: " .$sql. "</br>" .$conn->error; 
  }

  $conn->close();

  echo "Connected sucessfully.";
?>

Answer 1

當您的變量是數字時，您可以完全按照您對 id 所做的操作使用此代碼，如下所示：

$sql = "插入1 (chave,sensor1,sensor2,sensor3) 值 ($chave, $sensor1, $sensor2, $sensor3);";

但是當你的變量是文本時，例如你需要這個代碼：

$sql = "INSERT INTO 1 (chave,sensor1,sensor2,sensor3) 值 ('$chave', '$sensor1', '$sensor2', '$sensor3');";