Question

报错 Notice: Undefined index: id in D:\xampp\htdocs\web_health\main.php on line 65怎么改

<form id="form1" name="form1" method="GET" action= "<?php echo $_SERVER['PHP_SELF']?>" onsubmit="return Checker() " >
<p>
<label><input type="checkbox" name="item " value="personal_basic_information" id="chk_1" />基本信息表</label>

<label><input type="checkbox" name="item " value="principal_disease" id="chk_2" />重要疾病</label>

<label><input type="checkbox" name="item " value="baby" id="chk_3" />婴幼儿</label>

<label><input type="checkbox" name="item " value="pregnant_woman" id="chk_4" />孕产妇</label></p>
<input type="text" name="id" />
<p> <input name="查询提交按钮" type="submit" value="Submin" /><input type=reset value=重置></p>

</form>

<?php
$con = mysql_connect('localhost','root','');
if (!$con){
die('Could not connect: ' . mysql_error());
}else{
mysql_query("SET NAMES 'utf8'");
}
$identity_num = $_GET["id"];
$sql="SELECT * FROM web_health.personal_basic_information WHERE identity_num like('%"+".$identity_num."+"%')";
$query = mysql_query($sql);
echo "SELECT * FROM web_health.personal_basic_information WHERE identity_num like('%"+".$identity_num."+"%')";
echo $sql;
//echo $query;
$row=mysql_num_rows($query);
echo $row;
?>

Answer 1

改为这样就行了
$con = mysql_connect('localhost','root','');
if (!$con){
die('Could not connect: ' . mysql_error());
}else{
mysql_query("SET NAMES 'utf8'");
}
if(isset($_GET['id'])){
$identity_num = $_GET["id"];
$sql="SELECT * FROM web_health.personal_basic_information WHERE identity_num like('%"+".$identity_num."+"%')";
$query = mysql_query($sql);
echo "SELECT * FROM web_health.personal_basic_information WHERE identity_num like('%"+".$identity_num."+"%')";
echo $sql;
//echo $query;
$row=mysql_num_rows($query);
echo $row;
}
或者把这句$identity_num = $_GET["id"];改为$identity_num = @$_GET["id"];

追问

array(8) { [0]=> string(1) "1" ["id"]=> string(1) "1" [1]=> string(18) "220106199001089015" ["identity_num"]=> string(18) "220106199001089015" [2]=> string(6) "陈某某" ["name"]=> string(6) "陈某某" [3]=> string(2) "男" ["sex"]=> string(2) "男" } 请问为什么指出来第一条数据 并且重复了？

追答

mysql_num_rows只是输出行数，那
array(8) { [0]=> string(1) "1" ["id"]=> string(1) "1" [1]=> string(18) "220106199001089015" ["identity_num"]=> string(18) "220106199001089015" [2]=> string(6) "陈某某" ["name"]=> string(6) "陈某某" [3]=> string(2) "男" ["sex"]=> string(2) "男" }由什么函数输出的。你说清楚

追问

$row=mysql_num_rows($query);
echo $row;
就是这句输出的啊

追答

$row=mysql_num_rows($query);
echo $row;
这个$row输出来应该是数字呀。你可以百度一下这个函数。

追问

这是我输出的 可是在数据库中有好几个记录 只有第一个被输出了， 我用sql语句直接在数据库中查询是由四条记录