【发布时间】:2015-04-15 10:46:00
【问题描述】:
我有两张表,一张是变量,另一张是固定的。
$tab 结构
id int(11) autoincrement
cod text
den_material text
furnizor varchar(255)
cant_reper varchar(255)
lg varchar(255) No
cod_furnizor varchar(255)
obs tinytext
data date
和命名器表结构
nid int(11) autoincrement
sap text
denumire text
grupa varchar(255)
unitate
我想加入他们以在 $tab 中生成一个具有以下结构的单个表
id int(11) autoincrement
cod text --> to get datas from nomenclator table sap as column
den_material text -> to get datas from nomenclator table denumire as column
furnizor varchar(255)
cant_reper varchar(255)
lg varchar(255) No
cod_furnizor varchar(255)
obs tinytext
data date
完整的源代码
session_start();
// preserve selection for ajax call
if (!empty($_POST["tables"]))
{
$_SESSION["tab"] = $_POST["tables"];
$tab = $_SESSION["tab"];
}
// update on ajax call
if (!empty($_GET["grid_id"]))
$tab = $_SESSION["tab"];
if (!empty($tab))
{
$g = new jqgrid();
// set few params
$grid["caption"] = "Comanda : '$tab'";
$grid["autowidth"] = true;
$grid["multiselect"] = false; // allow you to multi-select through checkboxes
$grid["form"]["position"] = "center";
$grid["view_options"] = array("width"=>"500");
$g->select_command = "SELECT t.* FROM `$tab` AS t, `nomenclator` AS n
WHERE n.`sap` = t.`cod` OR n.`denumire` = t.`den_material`;";
$g->set_options($grid);
$g->set_actions(array(
"add"=>true, // allow/disallow add
"edit"=>true, // allow/disallow edit
"delete"=>true, // allow/disallow delete
"view"=>true, // allow/disallow delete
"rowactions"=>true, // show/hide row wise edit/del/save option
"search" => "advance", // show single/multi field search condition (e.g. simple or advance)
"showhidecolumns" => false
)
);
// set database table for CRUD operations
$g->table = $tab;
$col = array();
$col["title"] = "sap";
$col["name"] = "sap";
$col["width"] = "10";
$col["editable"] = true;
$col["hidden"] = true;
$cols[] = $col;
$col = array();
$col["index"] = "furnizor";
$col["title"] = "Comerciale asigurate de: ";
$col["editable"] = true;
$col["edittype"] = "select"; // render as checkbo
$col["editoptions"] = array("value"=>'Ramira:Ramira;Beneficiar:Beneficiar', "multiple" => true);
$cols[] = $col;
$col = array();
$col["index"] = "den_material";
$col["title"] = "Denumire Material";
$col["name"] = "denumire";
$col["editable"] = true;
$col["width"] = "80";
$col["searchoptions"] = array("value" => $str, "separator" => ":", "delimiter" => ";");
$col["search"] = true;
$col["formatter"] = "autocomplete"; // autocomplete
$col["formatoptions"] = array( "sql"=>"SELECT t.*
FROM `$tab` AS t, `nomenclator` AS n
WHERE n.`sap` = t.`cod` OR n.`denumire` = t.`den_material`;",
"search_on"=>"name",
"update_field" => "sap");
$cols[] = $col;
【问题讨论】:
-
你的意思是说你想将一个php变量加入到数据库中的一个表中?
-
你说“这会帮助 [你] 加薪”,所以我会建议你学习人们付钱给你做的工作。
-
@MaoTseTongue 说了什么。你所做的基本上是通过为你不知道的事情获得报酬来偷钱。学习技术,不要乞求代码sn-ps。如果你学不来,找一份适合你技能的工作。
-
投票结束,因为这是一个“为我编写代码”的问题。它也是任意数量的“我如何加入”问题的重复。
-
我不是程序员,我在物流部门工作,我们的 IT 不懂编程,我尝试了几次,但我做不到,这就是我寻求帮助的原因。
标签: php mysql sql database join