jQuery EasyUI - 添加指向单元格的链接 [英] jQuery EasyUI - Add link to cell

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问题描述

我有2个表:客户联系人。表联系人有一个指向客户表的外键。

I have 2 tables: Customer Contact. Table contact has a foreign key pointing at the customer table.

<table id="dgContactSearch" title="Suche" class="easyui-datagrid" style="height:160px" 
                            url="getContacts.php"
                            toolbar="#toolbarContactSearch" pagination="true"
                            rownumbers="true" fitColumns="true" singleSelect="true">
                        <thead>
                            <tr>
                                <th field="customer_id" width="50">Customer</th>
                                <th field="name" width="50">Name</th>
                                <th field="function" width="50">Funktion</th>
                                <th field="phone" width="50">Phone</th>
                                <th field="mobile" width="50">Mobile</th>
                                <th field="fax" width="50">Fax</th>
                                <th field="email" width="50">Email</th>
                                <th field="comment" width="50">Commnent</th>
                            </tr>
                        </thead>
                    </table>

这里是php文件

$page = isset($_POST['page']) ? intval($_POST['page']) : 1;
    $rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;
    $searchItemContact = isset($_POST['searchItemContact']) ? mysql_real_escape_string($_POST['searchItemContact']) : '';

    $searchItemContact = htmlentities($searchItemContact, ENT_QUOTES, 'UTF-8');

    $offset = ($page-1)*$rows;

    $result = array();

    $where = "name like '%$searchItemContact%' OR function like '%$searchItemContact%' OR email like '%$searchItemContact%' OR comment like '%$searchItemContact%'";
    $rs = mysql_query("select count(*) from contact where " . $where);
    $row = mysql_fetch_row($rs);
    $result["total"] = $row[0];

    $rs = mysql_query("select * from contact where " . $where . " limit $offset,$rows");

    $items = array();
    while($row = mysql_fetch_object($rs)){
        array_push($items, $row);
    }
    $result["rows"] = $items;
    echo json_encode($result);

我想要< th field =customer_idwidth = 50>客户< / th> 可链接。 ie < a href =customerView.php?id = $ customer_id>#< / a>
所以当我加载表格时,我看到了客户名称安装了id并链接到客户页面!
请帮忙。

I want to make the <th field="customer_id" width="50">Customer</th> linkable . i.e. <a href="customerView.php?id=$customer_id>#</a> So when I load the table I see the customer name instaed of the id and make a link to th e customer page! Please help.

更新

我以某种方式管理了一些解决方法:

I managed somehow a little workaround:

<th data-options="field:'name',width:100,align:'left',formatter:formatCustomerId">Name</th>

然后是Javascript函数:

and then Javascript function that goes along:

function formatCustomerId(val,row){
    var url = "customerView.php?id=";
    return '<a href="'+url + row.customer_id+'">'+val+'</a>';
}


推荐答案

最终解决方案是以前所有答案的混合:) 。

Final solution was a mixture of all the previous answers :).

<table id="dgContactSearch" title="Benutzer Suche" class="easyui-datagrid" style="height:160px" 
                            url="getContacts.php"
                            toolbar="#toolbarContactSearch" pagination="true"
                            rownumbers="true" fitColumns="true" singleSelect="true">
                        <thead>
                            <tr>
                                <th data-options="field:'firma',width:80,align:'left',formatter:formatCustomerId">Kunde</th>
                                <th data-options="field:'name',width:50,align:'left',formatter:formatContactUrl">Name</th>
                                <th field="function" width="50">Funktion</th>
                                <th field="phone" width="50">Phone</th>
                                <th field="email" width="50">Email</th>
                                <th field="mobile" width="50">Mobile</th>
                                <th field="fax" width="50">Fax</th>

                                <th field="comment" width="120">Kommentare</th>
                            </tr>
                        </thead>
</table>

剧本:

<script>
function formatCustomerId(val,row){
    var url = "customerView.php?id=";
    return '<a href="'+url + row.customer_id+'">'+val+'</a>';
}

function formatContactUrl(val,row){
    var url = "contactView.php?id=";
    return '<a href="'+url + row.id+'">'+val+'</a>';
}
</script>

和getContacts.php

and the getContacts.php

<?php
    include 'includes/db_functions.php';
    db_link();

    $page = isset($_POST['page']) ? intval($_POST['page']) : 1;
    $rows = isset($_POST['rows']) ? intval($_POST['rows']) : 10;
    $searchItemContact = isset($_POST['searchItemContact']) ? mysql_real_escape_string($_POST['searchItemContact']) : '';

    $searchItemContact = htmlentities($searchItemContact, ENT_QUOTES, 'UTF-8');

    $offset = ($page-1)*$rows;

    $result = array();

    $where = "name like '%$searchItemContact%' OR function like '%$searchItemContact%' OR customer.name like '%$searchItemContact%' OR email like '%$searchItemContact%' OR comment like '%$searchItemContact%'";
    $rs = mysql_query("select count(*) from contact where " . $where);
    $row = mysql_fetch_row($rs);
    $result["total"] = $row[0];


    $sql = "SELECT contact.*, customer.name
    FROM `contact`
    INNER JOIN `kunden` on contact.customer_id = kunden.id ";

    $rs = mysql_query($sql . "where " . $where . " limit $offset,$rows");

    $items = array();
    while($row = mysql_fetch_object($rs)){
        array_push($items, $row);
    }
    $result["rows"] = $items;
    echo json_encode($result);

?>

这篇关于jQuery EasyUI - 添加指向单元格的链接的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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