Question

我正在尝试从php中的形式获取两个值，即体重和性别使用jquery和ajax显示与该性别和体重相对应的价格

mysql表如下所示

gender range1  range2  price
male     0       100     20
male    101      200     30
female   0       100     25
female  101      199     46

php形式如下

  <input type="text" name="gender" id="gender">
  <input type="number" name="weight">
 price <span id="result"></span>
<input type="button" herf="javascript:;" onclick="getprice" value="">

ajax函数如下所示

    function getprice (gender, whieght) {
        var str_num {
        "gender" : gender,
        "weight" : weight
       };
             $.ajax({ data: str_num, 
        url: 'selectprice.php',
        type: 'post'
        beforeSend: function() {
     $("#result").html("in progress..");},
           succes: function (reponse){
         $("result").html(reponse);
                  })}}

selectprice.php看起来像这样

 $gender = $_GET["gender"];
  $weight = $_GET["weight"];
   $fetch = "SELECT * FROM table where gender like $gender and weight 
  >range1 and <range2"; 
$result = mysqli_query($con, $fetch) or die("Ocurrio un error en la 
  consulta SQL");
while ($row = $resul->fetch_assoc()) {
    echo "".$row["price"]."";
    }
   echo $result;

当我单击该按钮时，什么也没有发生，我真的可以使用一些帮助，在此先感谢

Answer 1

您真的把错误打包了。

<input type="text" name="gender" id="gender">
<input type="number" name="weight" id="weight">
price 
<input type="button" herf="javascript:;" onclick="getprice()" value="">


<script>
$(document).ready(function() {
});
function getprice () {
    var str_num = {
        "gender" : $('#gender').val(),
        "weight" : $('#weight').val()
    };
    $.ajax({ 
        data: str_num, 
        url: 'selectprice.php',
        type: 'post',
        beforeSend: function() {
            $("#result").html("in progress..");
        },
        succes: function (reponse){
            $("result").html(reponse);
        }
    })
};

</script>

PHP文件

<?php
    error_reporting(E_ALL);
    ini_set('display_errors', true);
//  ini_set('log_errors', true);

var_dump($_POST); // Your doing a post not a get in the javaacript.

Answer 2

您需要更新行：

url: 'selectprice.php?weight='+weight+'&gender='+gender,

或者您可以在data之后通过url

url: 'selectprice.php',
data: {weight:weight, gender,gender},

Answer 3

您在此处混合使用“ POST”和“ GET”。

如果您想使用“ GET”，请遵循Naveed的建议

如果要使用“ POST”，请在下面找到两种可能的解决方案：

1）简单的一种：将变量直接包含在数据中

 $gender = $_GET["gender"];  $weight = $_GET["weight"];

selectprice.php：

替换

 $gender = $_POST["gender"];  $weight = $_POST["weight"];

使用

function getprice (gender, weight) {
    var data={
    "gender" : gender,
    "weight" : weight
   };

var str_num = JSON.stringify(data);

         $.ajax({ data: str_num, 
    url: 'selectprice.php',
    type: 'post'
    beforeSend: function() {
 $("#result").html("in progress..");},
       success: function (reponse){
     $("result").html(reponse);
              })}}

2）稍微复杂一点：使用stringify

 $gender = $_GET["gender"];  $weight = $_GET["weight"];

selectprice.php：

替换

$json_data_from_js= json_decode($_POST['data']);
$gender = $json_data_from_js->gender ;
$weight = $json_data_from_js-> weight;

使用

scatter = plt.scatter(x, y)
ax = scatter.axes
ax.invert_xaxis()
ax.invert_yaxis()

如何获取两个值以获取结果在php和ajax中

3 个答案: