根据从数据库中选择的值搜索数据库 [英] search the database based on the value selected from database
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问题描述
有两个页面的级联drop下拉列表。首先是index.php，其次是fetch_state.php。那里的代码是
index.php
< body>
< div id =container>
< div id =body>
< div class =mhead>< h2>使用jQuery Ajax和PHP的级联下拉列表 - InfoTuts< / h2>< / div>
< form class =form-horizontalrole =formaction =enctype =multipart / form-datamethod =post>
< div id =dropdowns>
< div id =centerclass =cascade>
<？php
$ sql =SELECT * FROM search_parent ORDER BY searchname;
$ query = mysqli_query（$ con，$ sql）;
？>
< label>国家/地区：
< select name =countryid =drop1>
< option value =>请选择< / option>
<？php while（$ rs = mysqli_fetch_array（$ query，MYSQLI_ASSOC））{？>
< option value =<？php echo $ rs [id];？>><？php echo $ rs [searchname]; ？>< /选项>
<？php}？>
< / select>
< / label>
< / div>
< div class =cascadeid =state>< / div>
< div id =cityclass =cascade>< / div>
< / div>
< div class =col-md-4 col-sm-6>
< div class =media-body>
< div class =col-md-8>
< input class =btn btn-primaryvalue =Searchtype =submitname =submit>
< / div>
< / div>
< / div>
< / form>
< / div>
< / div>
< script src =jquery-1.11.1.js>< / script>
< script>
$（document）.ready（function（）
{
$（select＃drop1）。change（function（）{
var parent_id = $（select＃ （）;
// alert（parent_id）;
$（＃state）。html（）;
// $ （#city）。html（）;
if（parent_id.length> 0）{
$ .ajax（{
type：POST，
url：fetch_state.php，
data：parent_id =+ parent_id，
cache：false，
beforeSend：function（）{
$（'＃state' .html（'< img src =loader.gifalt =width =24height =24>'）;
}，
success：function（html）
$（＃state）。html（html）;
}
}）;
}
}）;
}）;
< / script>
< / body>
fetch_state.php
<？php
include（connection.php）;
$ parent_id = trim（mysql_escape_string（$ _ POST [parent_id]））;
$ sql =SELECT * FROM features WHERE parent_id =。$ parent_id。ORDER BY fname;
$ count = mysqli_num_rows（mysqli_query（$ con，$ sql））;
if（$ count> 0）{
$ query = mysqli_query（$ con，$ sql）;
？>
< label>状态：
< select name =stateid =drop2>
< option value =>请选择< / option>
<？php while（$ rs = mysqli_fetch_array（$ query，MYSQLI_ASSOC））{？>
< option value =<？php echo $ rs [id];？>><？php echo $ rs [fname]; ？>< /选项>
<？php}？>
< / select>
< / label>
<？php
}
？>
< script src =jquery-1.11.1.js>< / script>
上面的代码有助于我的级联列表中我希望执行的搜索代码
<？php
$ fname = mysqli_real_escape_string（$ con，$ _POST ['fname']）;
$ sql1 =SELECT * FROM office WHERE fname LIKE'％$ fname％';
$ result = mysqli_query（$ con，$ sql1）;
if（mysqli_num_rows（$ result）> 0）
{
while（$ row = mysqli_fetch_assoc（$ result））{
echo $ row [fname];
/ /想从表格中获取所有数据
}
} else {
echo0 results;
}
mysqli_close（$ con）;
？>
如前所述，当用户从第二个下拉列表我希望根据第二个下拉列表中选择的参数获取搜索结果，结果应显示在搜索栏下方的同一页面上
解决方案
你的HTML代码添加
< div class =showsearch> < / div>
js代码
$（document）.ready（function（）{
$（'＃drop2'）。on变化，函数（）{
变种FNAME = $（本）.VAL（）;
//重命名您的文件，其中包含$ fname与get_search_data.php
if（fname！==）{
$ .post（'get_search_data.php'，{fname：fname} ，function（data）{
$（'。showsearch'）。html（data）;
}）;
}
}）;
}）;
- 该代码后应显示$ row ['fname'] to showsearch div
- 你必须包含（connection.php）;在get_search_data.php中
get_search_data.php
<？php
include（connection.php）;
if（isset（$ _ POST ['fname']））{
$ fname = mysqli_real_escape_string（$ con，$ _POST ['fname']）;
}
$ sql1 ='SELECT * FROM office WHERE fname LIKE％'。$ fname。'％';
$ result = mysqli_query（$ con，$ sql1）;
if（mysqli_num_rows（$ result）> 0）
{
while（$ row = mysqli_fetch_assoc（$ result））{
echo $ row [fname ]; //这个数据应该出现在showsearch div
}
} else {
echo0 results;
}
mysqli_close（$ con）;
？>
I have a code for cascade drop down list that works perfectly fine, there are two drop down list and the 2nd one depends on the first one, however i wish to perform a search (from database table )based on the parameter that is choosed from the second drop down list. i have the code for search also, but i don't know how to combine the search code with the cascade drop down list code
There are two pages for the cascade drop down list. First is index.php and second is fetch_state.php. there code is
index.php
<body>
<div id="container">
<div id="body">
<div class="mhead"><h2>Cascaded dropdown with jQuery Ajax and PHP - InfoTuts</h2></div>
<form class="form-horizontal" role="form" action="" enctype="multipart/form-data" method="post">
<div id="dropdowns">
<div id="center" class="cascade">
<?php
$sql = "SELECT * FROM search_parent ORDER BY searchname";
$query = mysqli_query($con, $sql);
?>
<label>Country:
<select name="country" id = "drop1">
<option value="">Please Select</option>
<?php while ($rs = mysqli_fetch_array($query, MYSQLI_ASSOC )) { ?>
<option value="<?php echo $rs["id"]; ?>"><?php echo $rs["searchname"]; ?></option>
<?php } ?>
</select>
</label>
</div>
<div class="cascade" id="state"></div>
<div id="city" class="cascade"></div>
</div>
<div class="col-md-4 col-sm-6">
<div class="media-body">
<div class="col-md-8">
<input class="btn btn-primary" value="Search" type="submit" name="submit">
</div>
</div>
</div>
</form>
</div>
</div>
<script src="jquery-1.11.1.js"></script>
<script>
$(document).ready(function()
{
$("select#drop1").change(function(){
var parent_id = $("select#drop1 option:selected").attr('value');
// alert(parent_id);
$("#state").html( "" );
//$("#city").html( "" );
if (parent_id.length > 0 ){
$.ajax({
type: "POST",
url: "fetch_state.php",
data: "parent_id="+parent_id,
cache: false,
beforeSend: function (){
$('#state').html('<img src="loader.gif" alt="" width="24" height="24">');
},
success: function(html){
$("#state").html( html );
}
});
}
});
});
</script>
</body>
fetch_state.php
<?php
include("connection.php");
$parent_id = trim(mysql_escape_string($_POST["parent_id"]));
$sql = "SELECT * FROM features WHERE parent_id = ".$parent_id ." ORDER BY fname";
$count = mysqli_num_rows( mysqli_query($con, $sql) );
if ($count > 0 ) {
$query = mysqli_query($con, $sql);
?>
<label>State:
<select name="state" id="drop2">
<option value="">Please Select</option>
<?php while ($rs = mysqli_fetch_array($query, MYSQLI_ASSOC)) { ?>
<option value="<?php echo $rs["id"]; ?>"><?php echo $rs["fname"]; ?></option>
<?php } ?>
</select>
</label>
<?php
}
?>
<script src="jquery-1.11.1.js"></script>
the code above helps in my cascade list now the code for search that i wish to perform
<?php
$fname = mysqli_real_escape_string($con, $_POST['fname']);
$sql1 = "SELECT * FROM office WHERE fname LIKE '%$fname%';
$result = mysqli_query($con, $sql1);
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_assoc($result)) {
echo $row["fname"];
//would like to get all the data from the table here
}
} else {
echo "0 results";
}
mysqli_close($con);
?>
As i mentioned earlier also, when the user selects the value from the second dropdown list i wish to get the search result based on the parameter that is selected in the 2nd drop down list and the result should be displayed on the same page below the search bar
解决方案
after your html code add
<div class="showsearch"></div>
js code
$(document).ready(function(){
$('#drop2').on('change',function(){
var fname = $(this).val();
// rename your file which include $fname with get_search_data.php
if(fname !== ""){
$.post('get_search_data.php',{fname: fname},function(data){
$('.showsearch').html(data);
});
}
});
});
- after that code it should display the $row['fname'] to showsearch div
- you have to include("connection.php"); in get_search_data.php
get_search_data.php
<?php
include("connection.php");
if(isset($_POST['fname'])){
$fname = mysqli_real_escape_string($con, $_POST['fname']);
}
$sql1 = 'SELECT * FROM office WHERE fname LIKE "%'.$fname.'%"';
$result = mysqli_query($con, $sql1);
if (mysqli_num_rows($result) > 0)
{
while($row = mysqli_fetch_assoc($result)) {
echo $row["fname"]; // this data should appear in showsearch div
}
} else {
echo "0 results";
}
mysqli_close($con);
?>
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