php - 如何检查带参数的函数是否返回true

Question

这里我试图用 if function($String) == true if function returns true if returns true else 检查 if 条件 header('Location: home.php')我正在检查页面中的 if 条件，但看不到任何语句或页面重定向。

可能是什么问题？

这是我所做的。

<?php 
      if(ifproductPublished($qs) == true){
        echo "hey its true";
      } else {
        header('Location: index.php');
      }
?>

功能:

function ifproductPublished($productID){
    $conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    $sql = "SELECT product_status FROM product_list WHERE product_id='$productID'";
    $result = $conn->query($sql);
    if ($result->num_rows > 0) 
    {
        while($row = $result->fetch_assoc()) {      
        $product_status = $row["product_status"];                       
        if($product_status == 1){
            return true;
        } else {
            return false;
        }
        }       
        } else {
        echo "No results found";
    }
    $conn->close();
}

Answer 1

正如 @chris85 所提到的，您可以更改 sql，然后简化函数，例如:

function ifproductPublished( $productID ){
    $conn = new mysqli( DB_HOST, DB_USER, DB_PASS, DB_NAME );
    if ( $conn->connect_error ) die("Connection failed: " . $conn->connect_error);

    $sql = "SELECT `product_status` FROM `product_list` WHERE `product_id`='$productID' and `product_status`=1;";
    $result = $conn->query( $sql );
    $status = false;

    if ( $result->num_rows > 0 ) {                       
        $status=true;
    } else {
        echo "No results found";
    }
    $conn->close();
    return $status;
}