Question

我试图将数组数据插入数据库。有parent_array和child_arrays。

    ...
    titlePanel.setLayout(new GridBagLayout());
    titlePanel.setBackground(BLUE);

    header = new JLabel ("Gradebook");
    header.setLocation(200,400);
    header.setFont(new Font("Serif", Font.BOLD, 50));
    gbc.gridx = 0;
    gbc.gridy = 0;
    gbc.anchor = GridBagConstraints.CENTER;
    titlePanel.add(header,gbc);

    date = new Date();
    currentDate = new JLabel (fmt.format(date));
    currentDate.setFont(new Font("Serif", Font.PLAIN, 14));
    ActionListener updateTime = new ActionListener() {
        public void actionPerformed (ActionEvent e) {
            date = new Date();
            currentDate.setText(fmt.format(date));
        }
    };
    Timer timer = new Timer (1000, updateTime);
    timer.start();
    gbc.gridx = 0;
    gbc.gridy = 1;
    gbc.anchor = GridBagConstraints.CENTER;
    titlePanel.add(currentDate, gbc);

    JLabel userName = new JLabel ("Username:  ");
    gbc.gridx = 0;
    gbc.gridy = 2;
    gbc.anchor = GridBagConstraints.EAST;
    titlePanel.add(userName, gbc);

    JTextField username = new JTextField (10);
    gbc.gridx = 1;
    gbc.gridy = 2;
    gbc.anchor = GridBagConstraints.WEST;
    titlePanel.add(username, gbc);

    JLabel password = new JLabel ("Password:  ");
    gbc.gridx = 0;
    gbc.gridy = 3;
    gbc.anchor = GridBagConstraints.EAST;
    titlePanel.add(password, gbc);

    JPasswordField Password = new JPasswordField (10);
    gbc.gridx = 1;
    gbc.gridy = 3;
    gbc.anchor = GridBagConstraints.WEST;
    titlePanel.add(Password, gbc);
    JButton login = new JButton ("Login");
    gbc.gridx = 0;
    gbc.gridy = 4;
    gbc.anchor = GridBagConstraints.CENTER;
    titlePanel.add(login, gbc);

    JButton newAccount = new JButton ("Create New Account");
    gbc.gridx = 0;
    gbc.gridy = 5;
    gbc.anchor = GridBagConstraints.CENTER;
    titlePanel.add(newAccount, gbc);
    mainFrame.add(titlePanel);

HTML：

$arr_parent    = ['Apple','Banana','Grape'];        //parent_array
$arr_child_par = ['Price','Stock','Date_Stocked'];  //first child array ex.
$arr_child_val = ['1000','200','2012-05-05'];       //second child array ex.

现在，规则是这样的：

每个<input type='text' name='arr_parent[]' class='arr_parent' value='"+$('#fruit').val()+"' /> //Each table row has one this textbox <input type='text' name='arr_child_par[]' class='arr_child_par' value='Price' /> //Each table row has multiple of these textboxes <input type='text' name='arr_child_val[]' class='arr_child_val' value='1000' /> //Each table row has multiple of these textboxes长度和值都有所不同，具体取决于$arr_child_par
每个$arr_parent长度与$arr_child_val

最后，这是所需输出的表结构：

t_fruit

$arr_child_par

t_fruit_d

NO     FRUIT
 1      Apple
 2     Banana
 3      Grape

如何在NO FRUIT PARAMETER PARAMETER_VALUE 1 Apple Price 1000 2 Apple Stock 300 3 Banana Stock 400 4 Banana Supplier John 5 Banana Date_Stocked '2018-02-14' 6 Grape Price 3000中获取水果并在$arr_parent上使用它？

Answer 1

看起来你有3张桌子？果实，参数和参数值。因此，您应该使用外键，而不是使用参数的“标签” 所以它给出了类似的东西：

Table FRUIT
NO     label
 1      Apple
 2     Banana
 3      Grape

Table PARAM
NO     label    
 1      Price   
 2      Stock             
 3      Supplier
 4      Date_Stocked
 etc

Table PARAM_VAL
NO     FRUIT_ID      PARAM_ID      PARAM_VAL
 1     1             1             1000
 2     1             2             300
 3     2             2             400
 4     2             3             "John"

使用这种结构，你可以存储它并构建一个对象“fruit_d”但是这个解决方案在PHP中非常难看，并且可能对很多数据非常沉重......所以我建议使用SQL JOIN代替直接在php中构建你的fruit_d。

Answer 2

模式

CREATE TABLE fruits (
    fruit_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
    fruit VARCHAR(100) NOT NULL
    );

CREATE TABLE params (
    param_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
    fruit_id INT NOT NULL,
    FOREIGN KEY (fruit_id) REFERENCES fruit (fruit_id)
);

CREATE TABLE params_val (
    param_val_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
    fruit_id INT NOT NULL,
    param_id INT,
    FOREIGN KEY (fruit_id) REFERENCES fruits (fruit_id)
    FOREIGN KEY (param_id) REFERENCES params (param_id)
);

php更新并加入

<?php 
//connect and check
$con = mysqli_connect('localhost', 'root', '', 'mydb');
if (!$con) { exit(); }

// post and serialize
$fruit = implode("'","'", $_POST['arr_parent'] );
$param = implode("'","'", $_POST['arr_child_par'] );
$param_val = implode("'","'", $_POST['arr_child_val'] );

//db insert
$fruit_insert = "INSERT INTO fruits (label) VALUES ('.$fruit.')";
$param_insert = "INSERT INTO param (label) VALUES ('.$param.')";
$param_val_insert = "INSERT INTO param_val (label) VALUES ('.$param_val.')";

//join
$sql = "SELECT fruit, parameter, parameter_val
        FROM fruits
            LEFT OUTER JOIN params ON fruits.fruit_id = params.fruit_id
            INNER JOIN params_val ON params.param_id = params_val.param_id";

$result = mysql_query($sql);
?>

这就是想法......

如何在第二个循环中使用第一个循环的参数？

2 个答案: