如何从下拉列表

时间:2017-03-31 10:46:19

标签: php mysql select

我有一个Select标签,其中包含选项标签3的值 < option value ='"。$ row - > ownerID。","。$ row - > companyID。","。$ row - >。SITENAME"'>" $行级别> SITENAME"< /选项>&#34 ;; 例如:161,3,Burj El Brajneh 我需要在SQL查询中只比较第三个元素。 表: site_info: 的siteID 网站名 OWNERID companyID OWNER_INFO: OWNERID OWNERNAME company_info: companyID 公司名 选择标签:              < PHP                  $ query_site_name = $ wpdb-> get_results("从site_info&#34中选择DISTINCT siteNAME,ownerID,companyID;);                   foreach($ query_site_name为$ row)                   {                    echo"< option value ='"。$ row - > ownerID。","。$ row - > companyID。",&# 34;。$ row - > siteNAME。"'>"。$ row-> siteNAME。"< / option>&#34 ;;                   }              ?>             < /选择>< / TD> php检查: 如果(isset($ _ POST [' SITE_NAME']))           {            $ SITE_NAME = $ _ POST [' SITE_NAME'];           }           else {$ site_name ="&#34 ;; } 为了选择第三个元素,我希望我添加到SQL查询这个语句$ site_name-> siteNAME但它似乎不起作用 如果我在查询中只写$ site_name,结果将是: site_name =' 161,3,Burj El Brajneh' SQL查询:  $ sql = $ wpdb-> prepare("选择i.siteID          ,i.siteNAME          ,iequipmentTYPE          ,c.latitude          ,c.longitude          ,c。高度          ,o.ownerNAME          ,o.ownerCONTACT          ,x.companyNAME          ,y.subcontractorCOMPANY          ,y.subcontractorNAME          ,y.subcontractorCONTACT       来自site_info i       剩下       加入owner_info o         在i.ownerID = o.ownerID上       剩下       加入company_info x         在i.companyID = x.companyID上       剩下       JOIN subcontractor_info y         在i.subcontractorID = y.subcontractorID上         LEFT JOIN site_coordinates2 c         在i.siteID = c.siteID上         哪里         i.siteNAME =%s         和         o.ownerNAME =%s         和         x.companyNAME =%s        ",$ site_name-> SITENAME,$ OWNER_NAME,$ COMPANY_NAME);      $ query_submit = $ wpdb-> get_results($ sql,OBJECT);

1 个答案:

答案 0 :(得分:0)

在php

中这样
foreach (var item in source.GetItems().WithUsing())
{
    // ...
}

in html

IF ($row[$i]['abc'] == 'A'){
$a[$i]='<option value="A" selected="selected">A</option>';
$b[$i]='<option value="B" >B</option>';
}
IF ($row[$i]['abc'] == 'B'){
$a[$i]='<option value="A" >A</option>';
$b[$i]='<option value="B" selected="selected" B</option>';
}

您只需检查该值是否相同,然后将其回显为selected =“selected”

和你的if语句问题

<select>
    <php? echo $a[$i]; ?>
    <php? echo $b[$i]; ?>
 </select>