Javascript 使用ajax检查单选按钮时如何删除mysql?
使用ajax检查单选按钮时如何删除mysql 在我选中单选按钮后,第二个单选按钮 我想像这样删除mysqlJavascript 使用ajax检查单选按钮时如何删除mysql?,javascript,php,jquery,mysql,ajax,Javascript,Php,Jquery,Mysql,Ajax,使用ajax检查单选按钮时如何删除mysql 在我选中单选按钮后,第二个单选按钮 我想像这样删除mysql $sql = "DELETE FROM procust_store_1 WHERE products_id = '$_POST[products_id]' AND user = '$id' "; $objQuery = mysql_query($sql); $sql = "DELETE FROM procust_store_2 WHERE products_id = '$_POST
$sql = "DELETE FROM procust_store_1 WHERE products_id = '$_POST[products_id]' AND user = '$id' "; $objQuery = mysql_query($sql);
$sql = "DELETE FROM procust_store_2 WHERE products_id = '$_POST[products_id]' AND user = '$id' "; $objQuery = mysql_query($sql);
如果选中,则单选按钮第一个单选按钮
我想像这样删除mysql
$sql = "DELETE FROM procust_store_1 WHERE products_id = '$_POST[products_id]' AND user = '$id' "; $objQuery = mysql_query($sql);
$sql = "DELETE FROM procust_store_2 WHERE products_id = '$_POST[products_id]' AND user = '$id' "; $objQuery = mysql_query($sql);
我该怎么做
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
<script>
$(document).ready(function(){
$(".radio_id").click(function(){
if($(this).attr("value")=="first"){
$("#second_checkbox").hide();
$("#first_checkbox_display").show();
}
if($(this).attr("value")=="second"){
$("#first_checkbox_display").hide();
$("#second_checkbox").show();
}
});
});
</script>
<form id="first_checkbox_display" name="form1" method="post" action="" ENCTYPE = "multipart/form-data" onsubmit="return checkform(this);" style=" margin: 0px; " >
<input type="text" name="products_id" value="1294759">
<input type="text" name="products_color" value="red">
<input type="text" name="products_type" value="electronic">
<input type="button" value="Check1" onclick="doajax_products_check()"/>
<p id="myplace_data1"></p>
</form>
<form id="second_checkbox" name="form2" method="post" action="" ENCTYPE = "multipart/form-data" onsubmit="return checkform(this);" style=" display: none; margin: 0px; " >
<input type="text" name="products_id" value="0000000">
<input type="text" name="products_color" value="iiiiiii">
<input type="text" name="products_type" value="aaaaaaa">
<input type="button" value="Check2" onclick="doajax_products_check2()"/>
<p id="myplace_data2"></p>
</form>
<label><input type="radio" class="radio_id" name="colorRadio" value="first" checked >FIRST RADIO</label>
<br>
<label><input type="radio" class="radio_id" name="colorRadio" value="second">SECOND RADIO</label>
$(文档).ready(函数(){
$(“.radio_id”)。单击(函数(){
if($(this.attr(“value”)=“first”){
$(“#第二个#复选框”).hide();
$(“第一个复选框显示”).show();
}
if($(this.attr(“value”)=“second”){
$(“#第一个#复选框#显示”).hide();
$(“#第二个#复选框”).show();
}
});
});
第一台收音机
第二台收音机
mysql.*
已弃用。尝试使用mysqli.*
,您的查询易受sql注入攻击,vuln为:'或1=1/*