Ma puteti ajuta si pe mine la un insert !! Am urmatorul Script
Cod: Selectaţi tot
<script type="text/javascript">
function selectCity(country_id){
if(country_id!="-1"){
loadData('Familie',country_id);
$("#city_dropdown").html("<option value='-1'>Select city</option>");
}else{
$("#Familie_dropdown").html("<option value='-1'>Select state</option>");
$("#Produs_dropdown").html("<option value='-1'>Select city</option>");
}
}
function selectState(state_id){
if(state_id!="-1"){
loadData('Produs',state_id);
}else{
$("#Produs_dropdown").html("<option value='-1'>Select city</option>");
}
}
function loadData(loadType,loadId){
var dataString = 'loadType='+ loadType +'&loadId='+ loadId;
$("#"+loadType+"_loader").show();
$("#"+loadType+"_loader").fadeIn(400).html('Please wait... <img src="image/loading.gif" />');
$.ajax({
type: "POST",
url: "loadData.php",
data: dataString,
cache: false,
success: function(result){
$("#"+loadType+"_loader").hide();
$("#"+loadType+"_dropdown").html("<option value='-1'>"+loadType+"</option>");
$("#"+loadType+"_dropdown").append(result);
}
});
}
</script>Si partre de afisare care se alta tot in fisierul index.php
Cod: Selectaţi tot
<?php
if($checkCountry > 0){
?>
<form action='index.php' method='POST'>
<table>
<tr>
<td>
<select onchange="selectCity(this.options[this.selectedIndex].value)" name="client">
<option value="-1">Clientul</option>
<?php
while($rowCountry=mysql_fetch_array($resCountry)){
?>
<option value="<?php echo $rowCountry['id']?>"><?php echo $rowCountry['country_name']?></option>
<?php
}
?>
</select>
</td>
<td>
<select id="Familie_dropdown" onchange="selectState(this.options[this.selectedIndex].value)"name="familie">
<option value="-1">Familie Produs</option>
</select>
<span id="state_loader"></span>
</td>
<td>
<select id="Produs_dropdown" name="produs">
<option value="-1">Produs</option>
</select>
<span id="city_loader"></span>
</td>
<td>
<input type='text' name='cantitate'/>
</td>
<td>
<select name="formatie">
<option >Formatie</option>
<option>A</option>
<option>B</option>
<option>C</option>
</select>
</td>
<td>
<select name="schimb">
<option name="schimb">Schimb</option>
<option>1</option>
<option>2</option>
<option>3</option>
</select>
</td>
<td><input type='submit' value='Adauga' name='submit'/></td>
</tr>
</form>
</table>
<?php
}else{
echo 'No Country Name Found';
}Asa si avem si partea de inserare
Cod: Selectaţi tot
if(isset($_POST['submit'])){
$Client = $_POST['client'];
$familie= $_POST['familie'];
$Den = $_POST['produs'];
$cantitate = $_POST['cantitate'];
$Schimb = $_POST['schimb'];
$Formatia = $_POST['formatie'];
$sql4 = "insert into lista (Client,Produs,Cantitate,Schimb,Formatie,Data)
VALUES ('$Client','$Den','$cantitate','$Schimb','$Formatia',NOW())";
header('Location:index.php');
$res4 = mysql_query($sql4) or die(mysql_error());
}Problema este la inserare imi introduce la baza de date id fiecarui produs eu doresc denumirea ? se poate face asa ceva ? Va multumesc !!
P.S Fisierul loaddata.php
Cod: Selectaţi tot
<?php
include('dbConnect.inc.php');
$loadType=$_POST['loadType'];
$loadId=$_POST['loadId'];
if($loadType=="Familie"){
$sql="select id,state_name from state where country_id='".$loadId."' order by state_name asc";
}else{
$sql="select id,city_name from city where state_id='".$loadId."' order by city_name asc";
}
$res=mysql_query($sql);
$check=mysql_num_rows($res);
if($check > 0){
$HTML="";
while($row=mysql_fetch_array($res)){
$HTML.="<option value='".$row['id']."'>".$row['1']."</option>";
}
echo $HTML;
}
?>Multumesc incaodata !!
