从dropdownlist ini php的数据库onchange事件中恢复数据 [英] Retrive data from data base onchange event of dropdownlist ini php
本文介绍了从dropdownlist ini php的数据库onchange事件中恢复数据的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
< 表格 名称 = form1 方法 = post onsubmit =跨度> <跨度类= 代码关键字> xmlhttpPost( '为registration.php', 'form1中', 'colam', '请稍候');返回false;跨度> < span class =code-keyword>>
< table width = 600 align < span class =code-keyword> = center class = cellpadding = 1 cellspacing = 1 >
< tr >
< td align = right > N ame:< / td > < td align = left > < 输入 类型 = text 名称 = txt_name value = > < / td > < / tr >
< tr > < td align = right > 地址:< / td > < td 对齐 = left > < 输入 type = text name = txt_addrs value = > < / td > < / tr >
< tr > < td align = right > 电子邮件ID:< / td > < td > < 输入 type = text < span class =code-attribute> name = txt_email value = > < / td > < / tr >
< tr > < td align = right > 联系号:< / td > < td > < <跨度 class =code-leadattribute> input type = text name = txt_cnumber < span class =code-attribute> value = > < / td > < / tr >
< tr > < td align = 右 > 选择食物类型:< / td > < td > < select name = selectfood id = selectfood onchange = ajaxpage('choice.php?ID ='+ this.value,'choicefood') >
< span class =code-summarycomment><? php
< span class =code-summarycomment>
< span class =code-summarycomment> $ result = mysqli_query($ con, SELECT distinct item_type FROM < span class =code-summarycomment> tbl_master);
< span class =code-summarycomment> while($ row = mysqli_fetch_array($ result))
< span class =code-summarycomment> {? > ;
< 选项 value = <?php echo $ row ['item_type']; {echo 已选择;}? > > <? php echo ($ row ['item_type']);? > < / option >
<? php < span class =code-summarycomment>
< span class =code-summarycomment> < span class =code-summarycomment> }
? >
< / select >
< / td > < / tr >
< td colspan = 2 align = center >
< input type = submit name = 提交 value = 提交 > &安培; NBSP;&安培; NBSP;&安培; NBSP;&安培; NBSP;&安培; NBSP;
< 输入 < span class =code-attribute> type = hidden 名称 = 行动 value = ADD id = 行动 标题 = 提交 / >
< input 类型 = 重置 名称 = 重置 value = 重置 > < / td > < tr >
< / tr >
< / table >
< / form >
解决方案
结果 = mysqli_query(
con, SELECT distinct item_type FROM tbl_master);
< span class =code-summarycomment> while(
row = mysqli_fetch_array(
<form name="form1" method="post" onsubmit="xmlhttpPost('registration.php','form1','colam','please wait '); return false;">
<table width="600" align="center" class="" cellpadding="1" cellspacing="1">
<tr>
<td align="right">Name :</td><td align="left"><input type="text" name="txt_name" value=""></td></tr>
<tr><td align="right">Address :</td><td align="left"><input type="text" name="txt_addrs" value=""></td></tr>
<tr><td align="right">E- Mail ID :</td><td><input type="text" name="txt_email" value=""></td></tr>
<tr><td align="right">Contact No. :</td><td><input type="text" name="txt_cnumber" value=""></td></tr>
<tr><td align="right">Select Food Type :</td><td><select name="selectfood" id="selectfood" onchange="ajaxpage('choice.php?ID='+this.value,'choicefood')">
<?php
$result = mysqli_query($con,"SELECT distinct item_type FROM tbl_master");
while($row = mysqli_fetch_array($result))
{?>
<option value="<?php echo $row['item_type']; {echo "Selected";}?>"><?php echo ($row['item_type']);?></option>
<?php
}
?>
</select>
</td></tr>
<td colspan="2" align="center">
<input type="submit" name="submit" value="Submit">
<input type="hidden" name="ACTION" value="ADD" id="ACTION" title="submit" />
<input type="reset" name="reset" value="Reset"></td><tr>
</tr>
</table>
</form>
解决方案
result = mysqli_query(
con,"SELECT distinct item_type FROM tbl_master"); while(
row = mysqli_fetch_array(
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