Question

我查看了之前的所有示例，但仍然没有骰子，我有一个基本的php问题。

示例是here。我希望能够点击一个表格，其选项如下所示：

Example

在此代码中明确声明表名称有效：

if($_GET['action'] == 'getOptions'){
        $category = $_GET['category'];
        $query = "SELECT `COLUMN_NAME` 
                    FROM `INFORMATION_SCHEMA`.`COLUMNS` 
                    WHERE `TABLE_SCHEMA`='headfirstjson' AND 
                    **`TABLE_NAME`='AmericanOilProduction'**";

        $result = db_connection($query);
        //echo $result;
        $Options = array();

        while ($row = mysql_fetch_array($result)) {                
                $Options[] = $row;
        }
        echo json_encode(array("Options" => $Options));
        exit;
}

通过AJAX传递变量的这种组合不会：

AJAX：

function getOptions(category){
var category = category.value
$.ajax({
    url: "getData.php?action=getOptions",
    type: "GET",
    dataType:"json",
    data: {category:category},
    success: function(json){
        $.each(json.Options,function(){
        var option = "<option>"+this.COLUMN_NAME+"</option>"
        $('#options').append(option)
    });  
    }
});
}

PHP：

if($_GET['action'] == 'getOptions'){
        **$category = $_GET['category']**;
        $query = "SELECT `COLUMN_NAME` 
                    FROM `INFORMATION_SCHEMA`.`COLUMNS` 
                    WHERE `TABLE_SCHEMA`='headfirstjson' AND 
                    `TABLE_NAME`='**.$category.**'";

        $result = db_connection($query);
        //echo $result;
        $Options = array();

        while ($row = mysql_fetch_array($result)) {                
                $Options[] = $row;
        }
        echo json_encode(array("Options" => $Options));
        exit;
    }

如果有人可以帮助我，那就太好了！感谢。

Answer 1

您正在尝试将$category与其周围的句点连接起来。双引号将扩展变量，使表名为“.AmericanOilProduction。”。将代码更改为删除连接运算符。这里不需要它们。你也应该清理你的输入......

if($_GET['action'] == 'getOptions'){
    $category = $_GET['category'];
    $query = "SELECT `COLUMN_NAME` 
                FROM `INFORMATION_SCHEMA`.`COLUMNS` 
                WHERE `TABLE_SCHEMA`='headfirstjson' AND 
                `TABLE_NAME`='$category'";

    $result = db_connection($query);
    //echo $result;
    $Options = array();

    while ($row = mysql_fetch_array($result)) {                
            $Options[] = $row;
    }
    echo json_encode(array("Options" => $Options));
    exit;
}

Answer 2

我认为 jQuery因为URL本身和data:选项中存在的URL参数而感到困惑。

请尝试将action参数添加到data:选项中：

data: {
   category: category,
   action: 'getOptions'
},

将变量从Ajax传递给Php

2 个答案: