动态保存数据php

Question

我已将 ajax 数据与表单中的所有数据一起发布。用户可以为以前的雇主动态添加带有行的块。[empName]也许 10 或 20 ... 或 2 ...如何捕捉以前雇主的数量并将其保存到变量中？这是帖子数据：

["step-3"]=> string(1065) "{

"empYrXp":"min10yexp",

"empName-1":"TEST PAYPAL TEST",

"empStr-1":"TEST ORDER PAYPAL",

"empCity-1":"Pleven",

"empState-1":"CT",

"empContactPerson-1":"mbnmbnmbnm",

"empContactPersonPhone-1":"999-999-9999",

"empFromDate-1":"08-2013",

"empToDate-1":"06-2019",

"empPositionHeld-1":"bnmb",

"empReasonForLeaving-1":"nmbnmbnmbnm",

"empWereSubject-1":"No",

"empWasPrevJobDesignated-1":"No",

"empName-2":"TEST PAYPAL TEST",

"empStr-2":"TEST ORDER PAYPAL",

"empCity-2":"Pleven",

"empState-2":"CA",

"empContactPerson-2":"fghfhfhfgh",

"empContactPersonPhone-2":"999-999-9999",

"empFromDate-2":"02-2008",

"empToDate-2":"08-2013",

"empPositionHeld-2":"fghfgh",

"empReasonForLeaving-2":"fghfghfgh",

"empWereSubject-2":"No",

"empWasPrevJobDesignated-2":"No",

"empName-3":"TEST TEST",

"empStr-3":"TEST ORDER PAYPAL",

"empCity-3":"Pleven",

"empState-3":"CA",

"empContactPerson-3":"sdfsdfsdfsdf",

"empContactPersonPhone-3":"999-999-9999",

"empFromDate-3":"01-2006",

"empToDate-3":"01-2006",

"empPositionHeld-3":"sdfsdfs",

"empReasonForLeaving-3":"dfsdfsdfsdf",

"empWereSubject-3":"No",

"empWasPrevJobDesignated-3":"No"

}" 

我试着像这样计算以前的雇主

$countAllEmp = ( count($jsonDecode3) - 1 ) / 12;

Answer 1

不知道您如何获得 json，但表单支持命名，[]从而在服务器端生成数组：<input name="someName[]" /><input name="someName[]" /><input name="someName[]" />尝试输出$_POST['someName']，它将是具有 3 个元素的 0 索引数组。此外，您可以[]显式定义索引：<input name="someName[11]" /><input name="someName[12]" /><input name="someName[13]" />尝试输出$_POST['someName']，它将是具有 3 个键的数组 - 11, 12, 13。所以，你必须重新命名字段来代替empName-，empStr-等等-使用[]的符号：empName[]，empStr[]等等。