php - 如何使用 PDO 将 html 输入数组插入到 mysql 中

Question

我有一个 HTML 表单，其中有 2 个动态字段，它们是:

<input type="text"  name="ingredients[]" placeholder="Ingredients"/></div>
<input type="text"  name="quantity[]" placeholder="Quantity"/></div>

看看这里，对于表单中的输入，数据库表应该是这样的 英格 1 1001 1KG 英格2 1001 2KG 3 1001 3KG

现在检查图像，您就会看到发生了什么

DB Ingredients TABLE

我的 php 文件中有食谱 ID，我只需要一个 PDO 代码来帮助我将“成分”、“数量”和“$recipe_ID”插入一行。

<?php
header('Content-type: application/json');
require_once 'db/pdoconnect.php';
if($_POST)
    {

        $sth = $con->prepare("SELECT `recipeid` FROM Recipes ORDER BY `recipeid` DESC");
        $sth->execute();
        $previousid = $sth->fetchColumn();
        $offset=1;
        $generatingid=$previousid+$offset;
        $newid=$generatingid;
        //print("Last Id=$previousid\n");
        //print("New Id=$newid\n");

        $title = $_POST['title'];
        $preptime = $_POST['preptime'];
        $cocktime = $_POST['cocktime'];
        $level = $_POST['level'];
        $serving = $_POST['serving'];
        $recipetype = $_POST['recipetype'];
        $intro = $_POST['intro'];
        $details = $_POST['details'];
        $image = $_POST['image'];

        try
        {    

            $stmt = $con->prepare("INSERT INTO Recipes (recipeid,title,preptime,cocktime,level,serving,recipetype,intro,details,recipeimg) VALUES( :newid, :title, :preptime, :cocktime, :level, :serving, :recipetype, :intro, :details, :image)");
            $stmt->bindParam(":newid",$newid);
            $stmt->bindParam(":title",$title);
            $stmt->bindParam(":preptime",$preptime);
            $stmt->bindParam(":cocktime",$cocktime);
            $stmt->bindParam(":level",$level);
            $stmt->bindParam(":serving",$serving);
            $stmt->bindParam(":recipetype",$recipetype);
            $stmt->bindParam(":intro",$intro);
            $stmt->bindParam(":details",$details);
            $stmt->bindParam(":image",$image);

               if($stmt->execute()) { //check if main query has been executed 

    $sql = "INSERT INTO Ingredients VALUES";

    for($i = 1 ; $i <= count($_POST['ingredients']) ; $i++){
        $sql .= " (:recipeid" .$i. ", :ingredient" .$i. ", :quantity" .$i. "),"; 
    }

    // remove the last (,) from the $sql
    $sql = rtrim($sql, ',');
    $sth = $con->prepare($sql);

    // binding parameters 
    for($i = 1 ; $i <= count($_POST['ingredients']) ; $i++){
        $varIng = $_POST['ingredients'][$i];
        $varQnty = $_POST['quantity'][$i];

        $sth->bindParam(':recipeid' .$i , $newid , PDO::PARAM_STR);
        $sth->bindParam(':ingredient' .$i , $varIng , PDO::PARAM_STR);
        $sth->bindParam(':quantity' .$i , $varQnty , PDO::PARAM_STR);
    }  
              if ($sth->execute()) { 
              $response_array['status'] = 'success';  
                  }
              }//END OF FIRST IF STATEMENT

              else{
              $response_array['status'] = 'error';  
              }
               echo json_encode($response_array);   //SEND THE RESPONSE

               }//END OF TRY 

        catch(PDOException $e){
            echo $e->getMessage();
        }
    }

?>

Answer 1

假设您在数据库中的问题 ingredient_quantity 中发布的图片中显示了表格名称，并且您说您已经拥有 recipe-ID PHP Fiddle

<?php 
    $newid = 'A10';
    //considering this is your table shown in the picture
    $sql = "INSERT INTO ingredient_quantity VALUES";

    for($i = 1 ; $i <= count($_POST['ingredients']) ; $i++){
        $sql .= " (:newid" .$i. ", :ingredient" .$i. ", :quantity" .$i. "),"; 
    }

    // remove the last (,) from the $sql
    $sql = rtrim($sql, ',');
    $sth = $con->prepare($sql);

    // binding parameters 
    for($i = 1 ; $i <= count($_POST['ingredients']) ; $i++){
        $varIng = $_POST['ingredients'][$i];
        $varQnty = $_POST['quantity'][$i];
        $sth->bindParam(':newid' .$i , $newid , PDO::PARAM_STR);
        $sth->bindParam(':ingredient' .$i , $varIng , PDO::PARAM_STR);
        $sth->bindParam(':quantity' .$i , $varQnty , PDO::PARAM_STR);
    }
    $sth->execute();
?>

<小时/>

编辑1:

对于上面的代码，我有一个错误，因为ingredients[]数组从0开始，而不是1，最终的 for 循环的索引 将是未定义，因此解决这个问题，使最后一个 for 循环如下所示:

for($i = 0 ; $i < count($_POST['ingredients']) ; $i++){
        $varIng = $_POST['ingredients'][$i];
        $varQnty = $_POST['quantity'][$i];
        $j = $i + 1;
        $sth->bindParam(':newid' .$j , $newid , PDO::PARAM_STR);
        $sth->bindParam(':ingredient' .$j , $varIng , PDO::PARAM_STR);
        $sth->bindParam(':quantity' .$j , $varQnty , PDO::PARAM_STR);
    }

<小时/>

编辑2: 您可以尝试使用 ? 占位符而不是像这样的命名占位符 PHP Fiddle 2 :

//considering this is your table shown in the picture
    $sql = "INSERT INTO ingredient_quantity VALUES";

    for($i = 0 ; $i < count($_POST['ingredients']) ; $i++){
        $sql .= " ( ? , ? , ? ) , "; 
    }

    // remove the last (,) from the $sql
    $sql = rtrim($sql, ',');
    $sth = $con->prepare($sql);

    // binding parameters 
    $j = 1;
    for($i = 0 ; $i < count($_POST['ingredients']) ; $i++){
        $varIng = $_POST['ingredients'][$i];
        $varQnty = $_POST['quantity'][$i];
        $sth->bindValue( $j , $varIng);
        $sth->bindValue( $j + 1, $newid);
        $sth->bindValue( $j + 2, $varQnty);
        $j += 3;
    }
    $sth->execute();