我想問問東西..PHP - 如何插入jQuery的dropdown.val()到MySQL查詢
我已經陷在SQL查詢與jQuery這是dropdown.val()..
合併echo "<script type='text/javascript'>";
echo "$(document).ready(function() {
$('#ddlstatename').change(function(){";
echo "alert ($('#ddlstatename').val());";
$result6 = mysql_query("SELECT companyname FROM tb_company com inner join tb_companyinfo std on com.stateid = std.stateid and com.companyid = std.companyid WHERE std.productid = 1 AND std.stateid = '"$('#ddlstatename').val()"'");
echo "});";
echo "});</script>";
我儘量讓彈出:echo "alert ($('#ddlstatename').val());";
它從下拉菜單中(dllstatename是一個下拉ID)的值..
但是當我將其插入SQL查詢
$result6 = mysql_query("SELECT companyname FROM tb_company com inner join tb_companyinfo std on com.stateid = std.stateid and com.companyid = std.companyid WHERE std.productid = 1 AND std.stateid = '"$('#ddlstatename').val()"'");
我得到了錯誤:std.stateid = '"$('#ddlstatename').val()"'"..
如何改變$('#ddlstatename').val()
成sql查詢語句?
非常感謝您的幫助!
PHP的內容是服務器端,而JavaScript變量是客戶端。您必須使用表單發佈或ajax傳遞查詢的值 –
您提交表單! –