如何创建从 mysql 数据库中选择的新数组

Question

我正在尝试使用 PHP 在 mysql 数据库中将数据从一个表移动到另一个表。原始表具有一对多关系中的数据。有一列包含 project_id (value_object_id - 这是多列)，另一列 (value_field_id) 包含一个键，用于定义最后一列 (value_charvalue) 中存储的数据类型。我已经成功地能够使用以下代码提取所有数据，但无法弄清楚如何循环遍历每条记录以确定它是新的 project_id，在 value_field_id 列中包含 14、19 还是 20，以及如何将新记录的数据放入数组中。每条新记录都需要包含原始表中 3 条记录的信息。感谢您提供的任何帮助或指导。

<?php
$link = mysqli_connect("localhost", "username", "password", "test");

/* check connection */
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

$query = "SELECT value_object_id, value_field_id, value_charvalue FROM custom_fields_values ORDER BY value_object_id";

$result = mysqli_query($link, $query);

//  Add array for newly selected data????
//  $new_array = array();

/* numeric array */
while($row = mysqli_fetch_array($result, MYSQLI_NUM)) { 
    // Add test to determine if value_field_id is = 14, 19 or 20
    // $new_array[project_id] = value_object_id
    // If value_field_id = 19, $new_array[name] = value_charvalue
    // If value_field_id = 14, $new_array[email] = value_charvalue
    // If value_field_id = 20, $new_array[phone] = value_charvalue

    $project_id = $row[0];
    $next_row = $row[1];    
    $third_row = $row[2];
    printf('%s, %s, %s \n', $project_id, $next_row, $third_row);
    } 

/* free result set */
mysqli_free_result($result);

/* close connection */
mysqli_close($link);

?>

Answer 1

使用WHERE value_field_id IN (14, 19, 20)

$query = "SELECT value_object_id, value_field_id, value_charvalue FROM custom_fields_values ORDER BY value_object_id";

变成

$query = "SELECT value_object_id, value_field_id, value_charvalue FROM custom_fields_values WHERE value_field_id IN (14, 19, 20) ORDER BY value_object_id";

并操作您想要的任何内容或直接插入到其他表中。

Answer 2

所以首先您需要确定 value_field_id 是 19、14 还是 20。

// The field id's that you are interested in
$value_field_ids = array(19, 14, 20);

$project_id = array();

// I'm not sure why you would prefer numeric ids instead of associative?
while($row = mysqli_fetch_array($result, MYSQLI_NUM)) {

    // Here we check if the value_field_id is 19, 14 or 20
    if(in_array($row[1], $value_field_ids)) {
        $project_id[] = array($row);
    }

    $project_id = $row[0];
    $next_row = $row[1];  
    $third_row = $row[2];   
    printf('%s, %s, %s \n', $project_id, $next_row, $third_row);
}   

// At this point you have the entire data stored in $result, as an array
// Try var_dump($result);

你有了它——你新创建的数组。

Answer 3

<?php

$projects = [];

while(...) {
  ...

  $fieldname = ...;
  $fieldvalue = ...;

  if (!isset($projects[$project_id]))
    $projects[$project_id] = [];
  $projects[$project_id][$fieldname] = $fieldvalue;
}

print_r($projects);

如果您使用的是旧版本的 PHP，请将 [] 替换为 array()