这是我的HTML:
<div id="extraPerson">
<div class="controls controls-row">
<input class="span3" placeholder="First Name" type="text" name="firstname2">
<input class="span3" placeholder="Last Name" type="text" name="lastname2">
<select class="span2" name="gender2"><option value="Male">Male</option><option value="Female">Female</option></select>
...ETC
</div>
</div>
<a href="#" id="addRow"><i class="icon-plus-sign icon-white"></i> Add another family member</p></a>
这是我的jquery:
<script>
$(document).ready(function(){
$('#addRow').click(function(){
$("#extraPerson").slideDown();
});
})
</script>
#extraPerson隐藏在css中.
它会在单击链接时添加div.但是,我希望每次单击链接时继续添加相同的div.我该怎么做呢？如果将数字附加到表单输入名称,则更好.例如firstname3,firstname4等.
我需要格式化ISO 8601格式的时间戳(例如2001-10-26T21:32:52).当我date()在PHP中使用该函数时,它将替换T为Timezone(正如它应该做的那样).
我正在使用的命令是:
$time = date("y-m-dTH:i:s", time());
这会产生: 10-02-13EST10:21:03
如何让它插入实际T而不是替换EST？
我已经向REST API发出了oauth签名请求,并在数组中包含响应头,如下所示:
[0] => HTTP/1.1 200 OK
[1] => Cache-Control: private
[2] => Transfer-Encoding: chunked
[3] => Content-Type: text/html; charset=utf-8
[4] => Content-Location: https://***
[5] => Server: Microsoft-IIS/7.0
[6] => Set-Cookie: ASP.NET_SessionId=***; path=/; HttpOnly
[7] => X-AspNetMvc-Version: 2.0
[8] => oauth_token: ***
[9] => oauth_token_secret: ***
[10] => X-AspNet-Version: 4.0.30319
[11] => X-Powered-By: ASP.NET
[12] => Date: Sat, 15 Sep 2012 02:01:15 GMT
我试图找出如何解析标题以便轻松检索项目,如HTTP状态代码,Content-Location,oauth_token和oauth_token_secret？
我有以下json字符串,我想从中检索电子邮件地址.我怎么在PHP中做到这一点？
{"communications":{"communication":[{"@array":"true","@id":"23101384","@uri":"xyz/v1/Communications/1111","household":{"@id":"111111","@uri":"xyz/v1/Households/5465465"},"person":{"@id":"","@uri":""},"communicationType":{"@id":"1","@uri":"xyz/v1/Communications/CommunicationTypes/1","name":"Home Phone"},"communicationGeneralType":"Telephone","communicationValue":"1111","searchCommunicationValue":"2693240758","listed":"true","communicationComment":null,"createdDate":"2008-11-10T12:31:26","lastUpdatedDate":"2009-08-11T23:40:02"},{"@array":"true","@id":"11111","@uri":"xyz/v1/Communications/111111111","household":{"@id":"14436295","@uri":"xyz/v1/Households/11111"},"person":{"@id":"2222222","@uri":"xyz/v1/People/22222222"},"communicationType":{"@id":"2","@uri":"xyz/v1/Communications/CommunicationTypes/2","name":"Work Phone"},"communicationGeneralType":"Telephone","communicationValue":"11111","searchCommunicationValue":"789787987","listed":"false","communicationComment":null,"createdDate":"2009-08-09T15:49:27","lastUpdatedDate":"2009-08-11T23:40:02"},{"@array":"true","@id":"11111","@uri":"xyz/v1/Communications/11111","household":{"@id":"1111","@uri":"xyz/v1/Households/1111"},"person":{"@id":"244404","@uri":"xyz/v1/People/1111"},"communicationType":{"@id":"3","@uri":"xyz/v1/Communications/CommunicationTypes/3","name":"Mobile"},"communicationGeneralType":"Telephone","communicationValue":"22222","searchCommunicationValue":"5475454","listed":"true","communicationComment":null,"createdDate":"2008-11-10T12:31:26","lastUpdatedDate":"2009-08-11T23:40:02"},{"@array":"true","@id":"15454","@uri":"xyz/v1/Communications/111111","household":{"@id":"14436295","@uri":"xyz/v1/Households/1111"},"person":{"@id":"244444474","@uri":"xyz/v1/People/111111"},"communicationType":{"@id":"4","@uri":"xyz/v1/Communications/CommunicationTypes/4","name":"Email"},"communicationGeneralType":"Email","communicationValue":"email@needthis.com","searchCommunicationValue":"email@needthis.com","listed":"true","communicationComment":null,"createdDate":"2008-11-10T12:31:26","lastUpdatedDate":"2009-08-11T23:39:06"}]}}
我有一个看起来像这样的JSON字符串:
{"addresses":{"address":[{"@array":"true","@id":"888888","@uri":"xyz","household":{"@id":"44444","@uri":"xyz"},"person":{"@id":"","@uri":""},"addressType":{"@id":"1","@uri":"xyz","name":"Primary"},"address1":"xyz","address2":null,"address3":null,"city":"xyz","postalCode":"111111"}]}}
什么是解码此和地点PHP address1,address2,address3,city,和postalCode到会话变量？
到目前为止,我试过这个,但它不起作用:
$results = json_decode(strstr($address, '{"addresses":{"address":[{'), true);
$_SESSION['address1'] = $results['address']['address1'];
谢谢!
我有一个包含从javascript生成的值的html表.如何将这些值传递给php变量？
如何检查api响应中每个值的数据类型不是整数？例如，如果我的api返回以下内容：
"teamPermissions": [
"Edit",
"Administrator",
"ReadOnly",
etc
]
我需要进行邮递员测试，以确保在teamPermission数组中永不返回整数。这是我开始但需要帮助的内容：
var jsonData = JSON.parse(responseBody);
tests["Team Permissions Do Not Contain Integers"] = typeof(jsonData.teamPermissions) !== "number"
这样通过是因为teamPermissions是一个对象，但是如何检查该对象的每个值不是整数？
我试图弄清楚如何迭代一个解码的json字符串数组.我想为firstname,lastname等创建变量,然后将它们输出到表中的单独行.这是一个示例数组.大多数结果都会有多条记录.任何建议,将不胜感激:
Array ( [results] => Array ( [@count] => 1 [@pageNumber] => 1 [@totalRecords] => 1 [@additionalPages] => 0 [person] => Array ( [0] => Array ( [@array] => true [@id] => 38903211 [@uri] => https://api-name-removed.com/People/38903211 [@imageURI] => [@oldID] => 38614423 [@iCode] => fwVVyUOWEg3DOjIfg== [@householdID] => 23902641 [@oldHouseholdID] => 23740508 [title] => [salutation] => [prefix] => [firstName] => Mandy [lastName] => Ford [suffix] => [middleName] => [goesByName] => [formerName] => [gender] => Female [dateOfBirth] => 1965-02-11T00:00:00 [maritalStatus] => [householdMemberType] => …我有许多数据驱动的测试，因此我可以对多行数据运行相同的测试，效果很好。但是，我们还使用TestRail，并通过RF测试上的标签将RF测试链接到TestRail。目前，我只为每个模板标记一个TestRailID。例如：
*** Test Cases ***
Verify Registering For An Event with each CC Type
[Template] Register For An Event with a Credit Card
[Tags] TestRailID=1211 Smoke
${cc_intl} ${personInfo} ${visaCardInfo}
${cc_intl} ${personInfo} ${masterCardInfo}
${cc_intl} ${personInfo} ${americanCardInfo}
#etc
我希望每一行数据都具有TestRailID的唯一标记。在上面的示例中，如何为每个数据行添加标签？
我正在使用REST API将属性POST到使用json的人。我的请求正文如下所示：
$requestBody = '
{
"attribute": {
"@id": "",
"@uri": "",
"person": {
"@id": "222",
"@uri": "https://api_name_removed.com/v1/People/222"
},
"attributeGroup": {
"@id": "",
"@uri": "",
"name": null,
"attribute": {
"@id": "2404",
"@uri": "",
"name": null
}
},
"lastUpdatedDate": null
}
}';
如何将人员ID，人员uri和属性ID更改为已经存储的变量？
Noob问题.我有这个:
Array
(
[0] => address = 123 Something Street
[1] => address2 = Something else
[2] => city = Kalamazoo
[3] => state = MI
[4] => zip = 49097
[5] => country = United States
)
但我想要这个:
Array
(
[address] => 123 Something Street
[address2] => Something else
[city] => Kalamazoo
[state] => MI
[zip] => 49097
[country] => United States
)
我该怎么做呢？谢谢!
我有这样的模型类:
public class TypeMinimal : BaseMinimal {
private string _name = "";
public TypeMinimal(int? id, string name = "")
: base(id, "types", API.Enum.Types) {
_name = name;
}
[DataMember]
public string name { get { return _name; } set { _name = value; } }
}
我试图根据类型是1还是2来填充属性.例如
type = new TypeMinimal(item.Type == 1 ? 1 : 2)
我想将类型名称和类型ID添加到属性,但我不确定语法.这是我尝试过的但是不正确:
type = new TypeMinimal(item.Type == 1 ? 1, "volunteer" : 2, "staff"),
这样做的正确方法是什么？