[英]In Ajax, How can I get 2 input values by the script code?
I am making the 2 input values to search my database in detail. 我正在输入2个输入值来详细搜索数据库。 But in ajax, the script code is good at 1 input value, but poor at 2 input values.
但是在ajax中,脚本代码在1个输入值时好,但在2个输入值时差。 So I have show you my codes as like below.
因此,我向您展示了我的代码,如下所示。 First, My html code and script code.
首先,我的html代码和脚本代码。
<script>
function showUser1(str)
{
----
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint10").innerHTML=xmlhttp.responseText;
}
}
var lang20 = document.getElementById('lang20').value;
xmlhttp.open("GET","./database/db1_" +lang20 + ".php?q="+qstr+ "&p="+pstr,true);
xmlhttp.send();
}
</script>
<form>
<select name="lang20" id="lang20" title="choose the language you want">
<option value="co">한국어</option>
<option value="en">English</option>
<option value="af">Afrikaans</option>
</select>
Input: <input name="search" onkeyup="showUser1(this.value)" >
Input: <input name="search" onkeyup="showUser1(this.value)" >
</form>
<div id="txtHint10"><b>Disease information will be listed here.</b></div>
Second, my db1_co.php code is as like below. 其次,我的db1_co.php代码如下所示。
<?php
$q = htmlspecialchars($_GET['q']);
$p = htmlspecialchars($_GET['p']);
$con = mysqli_connect('localhost',---);
if (!$con)
{
---
}
mysqli_select_db($con,"ajax_demo");
$sql = "SELECT code_co.code, code_co.disease_co, note.note, inclusion.inclusion,
advertiser.drug, subject.subject, subject.icd_category FROM code_co left join subject
on subject.code=code_co.code left JOIN note ON code_co.code = note.code left JOIN
inclusion ON code_co.code = inclusion.code left JOIN advertiser ON code_co.code =
advertiser.code WHERE code_co.disease_co LIKE '%".$q."%' and code_co.disease_co LIKE
'%".$p."%' OR code_co.code like '%".$q."%' and code_co.code like '%".$p."%'" ;
Above, I guess my script may be wrong. 以上,我想我的脚本可能是错误的。 That is, 'php?q="+qstr+ "&p="+pstr,true);'
也就是说,'php?q =“ + qstr +”&p =“ + pstr,true);' is wrong. Please help me. Give me a piece of advice, please. Thank you for your concern.
是错的,请帮助我,请给我一些建议,谢谢您的关注。
Change name of other input, for example name="search2"
and giv ID to them too. 也更改其他输入的名称,例如
name="search2"
和giv ID。
<input name="search" id="srch1" onkeyup="showUser1(this.value)" />
<input name="search2" id="srch2" onkeyup="showUser1(this.value)" />
var lang20 = document.getElementById('lang20').value;
var srch1 = document.getElementById('srch1').value;
var srch2 = document.getElementById('srch2').value;
xmlhttp.open("GET","./database/db1_" +lang20 + ".php?srch1="+srch1+ "&srch2="+srch2,true);
xmlhttp.send();
And then in PHP get values as this, 然后在PHP中获取这样的值,
$srch1 = $_GET['srch1'];
$srch2 = $_GET['srch2'];
you can try this 你可以试试这个
SCRIPT: 脚本:
<script>
function showUser1() // call function without parameter
{
if (window.XMLHttpRequest)
{
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{
// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint10").innerHTML=xmlhttp.responseText;
}
}
var lang20 = document.getElementById('lang20').value;
var qstr = document.getElementById('qstr').value; // get value of qstr
var pstr = document.getElementById('pstr').value; // get value of pstr
xmlhttp.open("GET","./database/db1_" +lang20 + ".php?q="+qstr+ "&p="+pstr,true);
xmlhttp.send();
}
</script>
HTML : HTML:
<form>
<select name="lang20" id="lang20" title="choose the language you want">
<option value="co">한국어</option>
<option value="en">English</option>
<option value="af">Afrikaans</option>
</select>
Input: <input name="search" id="qstr" onkeyup="showUser1()" > <!-- give id to qstr and call function withoud this.value -->
Input: <input name="search" id="pstr" onkeyup="showUser1()" > <!-- give id to pstr and call function withoud this.value -->
</form>
<div id="txtHint10"><b>Disease information will be listed here.</b></div>
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