Question

我正在尝试使用AJAX获取PHP脚本并将值作为JSON返回。出于某种原因，我的脚本失败了，我试图找出问题所在。当我从数据库中输入一个值到地址栏时，如：

www.someaddress/post.php?kinaseEntry=aValue

我得到一个JSON输出：

{"kinaseSKU":null,"url":null,"molecularWeight":null,"tracerSKU":null,"antiSKU1":"antiSKU1","antiSKU2":"antiSKU2","bufferSKU":"bufferSKU","tracerConc":null,"assayConc":null}

我的PHP文件如下：

<?php

//Include connection to database
require_once 'connect.php';

$kinase = mysql_real_escape_string ($_POST["kinaseEntry"]);

mysql_query('SET CHARACTER SET utf8'); 

$findKinase = "SELECT * FROM kbaData where cleanSKU = '" .$kinase. "' ";

if ($result = mysql_query($findKinase)) {

    $row = mysql_fetch_array($result, MYSQL_ASSOC);

    $kinaseSKU = $row['cleanSKU'];
    $url = $row['url'];
    $molecularWeight = $row['molecularWeight'];
    $tracerSKU = $row['tracerSKU'];
    $antiSKU1 = $row['antiSKU1'];
    $antiSKU2 = $row['antiSKU2'];
    $bufferSKU = $row['bufferSKU'];
    $tracerConc = $row['tracerConc'];
    $assayConc = $row['assayConc'];


    /* JSON ROW */
    $json = array ("kinaseSKU" => $kinaseSKU, "url" => $url, "molecularWeight" => $molecularWeight, "tracerSKU" => $tracerSKU, "antiSKU1" => $antiSKU1, "antiSKU2" => $antiSKU2, "bufferSKU" => $bufferSKU, "tracerConc" => $tracerConc, "assayConc" => $assayConc );

} else  {

    /* CATCH ANY ERRORS */
    $json = array('error' => 'Mysql Query Error');
}

/* SEND AS JSON */
header("Content-Type: application/json", true);

/* RETURN JSON */
echo json_encode($json);

/* STOP SCRIPT */
exit;
?>

我是以错误的方式来做这件事的吗？或者我做错了吗？

编辑：这是我调用PHP脚本的jQuery / Ajax：

$(document).ready(function()    {

$('#kinaseEntry').change(function ()    {

    var kinaseEntry = $('#kinaseEntry').val();
    var dataString = 'kinaseEntry' + kinaseEntry;

    $('#waiting').show(500);
    $('#message').hide(0);
    alert(kinaseEntry);

    //Fetch list from database
    $.ajax({
        type : "POST",
        url : "post.php",
        datatype: "json",
        data:   dataString,
        success : function(datas)   {
            alert("datas" + datas);
        },
        error : function(error) {
            alert("Oops, there was an error!");
        }
    });
    return false;
});
});

Answer 1

即使返回0行，mysql_query函数也会返回true值。仅在出现数据库错误时才返回false。所以我相信它执行一个查询并且没有结果，因此JSON中有null个值。尝试使用mysql_num_rows：

$findKinase = "SELECT * FROM kbaData where cleanSKU = '" .$kinase. "' ";

if ($result = mysql_query($findKinase)) {
      $json = array('error' => 'Mysql Query Error');
}else{
    $num_rows = mysql_num_rows($result);
    if($num_rows > 0){
         // Do sth with the results
    }else{
         $json = array('error' => 'No results');
    }
}

Answer 2

您的问题似乎是您的查询实际上没有返回任何行。尝试：

$result = mysql_query($findKinase);
if ($result !== false && mysql_num_rows($result) > 0) {
    // your code
}

如果查询中存在错误，则

$result在此情况下为false。如果没有错误，则检查是否确实返回了行。如果超过0行返回，则尝试提取数据。

Answer 3

首先，我不确定if (isset($_POST['kinaseEntry']))是否适用于您所展示的内容。您显示的网址是获取请求，因此如果您想要访问该变量，则必须使用$_GET['kinaseEntry']。如果您需要进行POST，请将表单中的method属性更改为<form method="POST">，以便为您提供post变量。

Answer 4

是的，你回来了一系列行，所以这并不像你期望的那样真正起作用。这应该解决它，但我不认为这是最好的方法（但我认为它应该工作）。

$kinase = mysql_real_escape_string ($_POST["kinaseEntry"]);

mysql_query('SET CHARACTER SET utf8'); 

$findKinase = "SELECT * FROM kbaData where cleanSKU = '" .$kinase. "' , LIMIT 0,1";

if ($result = mysql_query($findKinase)) {

    $row = mysql_fetch_array($result, MYSQL_ASSOC);

    $kinaseSKU = $row[0]['cleanSKU'];
    $url = $row[0]['url'];
    $molecularWeight = $row[0]['molecularWeight'];
    $tracerSKU = $row[0]['tracerSKU'];
    $antiSKU1 = $row[0]['antiSKU1'];
    $antiSKU2 = $row[0]['antiSKU2'];
    $bufferSKU = $row[0]['bufferSKU'];
    $tracerConc = $row[0]['tracerConc'];
    $assayConc = $row[0]['assayConc'];


    /* JSON ROW */
    $json = array ("kinaseSKU" => $kinaseSKU, "url" => $url, "molecularWeight" => $molecularWeight, "tracerSKU" => $tracerSKU, "antiSKU1" => $antiSKU1, "antiSKU2" => $antiSKU2, "bufferSKU" => $bufferSKU, "tracerConc" => $tracerConc, "assayConc" => $assayConc );

} else  {

    /* CATCH ANY ERRORS */
    $json = array('error' => 'Mysql Query Error');
}

/* SEND AS JSON */
header("Content-Type: application/json", true);

/* RETURN JSON */
echo json_encode($json);

/* STOP SCRIPT */
exit;

JSON PHP数据返回null

4 个答案: