本文介绍了如何将JSON传递给复选框的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
Json
{segment:IV级学费,VI-VIII级学费
, Board:[cbse]
,Class IV:[Allsubject]
,Class VI-VIII:[cbse,cse / Ise]
,Class VI-VIII Subject:[Allsubject,Science]
,Class XI-X:null
,Class XI-X Subject:null
, Class XI-XII:null
,Class XI-XII Subject:null
,Languagesubject:null
,engineering:null
, :null
,Dancesubject:null
,Degree Course:null
,Degree Subject:null
}
这里的键: Board
value:cbse
key:Class IV
value:Allsubject like wise json pattern in classconducted column
我的问题是:
- 如何检索json字符串?
-
我试过这样。
$ sqledit = mysql_query(select * from tinfo where tsname = '。$ _ SESSION ['tutorname']。');
$ row = mysql_fetch_array($ sqledit);
foreach($ key => $ val){
if($ key!=''&& $ val!='' ){
echo'< span class =details>'。$ key。'< / span>'。':'。< br />< br />;
if(is_array($ val))
echo implode(',',$ val)。 <峰; br /><峰; br /> 中;
else
echo $ val。 <峰; br /> 中;
code $
现在我想编辑页面。所以使用
foreach
我分开这些值。这里我解码了json值。现在价值即将到来。我想知道如何将json值传递给我的表单。这是我的表格。我已经试过了,我不知道我的代码有什么问题。
我想要这样做:检查这是这个正确
< input type =checkboxname =board []value =cbse<?php if ($ val [0] =='cbse'){echochecked;}?>> CBSE
< input type =checkboxname =board []value =cse / Ise <?php if($ val [1] =='ICSE / ISE'){echochecked;}?>>> ICSE / ISE。
解决方案你的行为与,但解析JSON而不是爆炸它是。
$ cats = json_decode($ row ['classconducted'],true);
< input type =checkboxname =segment []value =Class IV Tuition<?php if(in_array('Class IV Tuition',$ cats)){echo 检查; }>?> I-V类学费
Json
{"segment":"Class I-V Tuition,Class VI-VIII Tuition" ,"Board":["cbse"] ,"Class I-V":["Allsubject"] ,"Class VI-VIII":["cbse","cse/Ise"] ,"Class VI-VIII Subject":["Allsubject","Science"] ,"Class XI-X":null ,"Class XI-X Subject":null ,"Class XI-XII":null ,"Class XI-XII Subject":null ,"Languagesubject":null ,"engineering":null ,"Diploma":null ,"Dancesubject":null ,"Degree Course":null ,"Degree Subject":null } Here key:Board value:cbse key:Class I-V value:Allsubject like wise json pattern stored in classconducted column
My questions are:
- How to retrieve the json string?
I have tried like this.
$sqledit = mysql_query("select * from tinfo where tsname='" .$_SESSION['tutorname'] ."'"); $row = mysql_fetch_array($sqledit); foreach (json_decode($row['classconducted'], true) as $key => $val) { if($key!='' && $val!='') { echo '<span class="details">'.$key.'</span>'.': '."<br/><br/>"; if (is_array($val)) echo implode(',', $val) . "<br/><br/>"; else echo $val . "<br/>"; } }
Now I want to edit the page. So using
foreach
I separate the values. Here I have decode the json values. now the values are coming . I want to know how to pass the json values to my form. Here is my form. I have tried like this, I don't know what is wrong in my code.I want ouptut like this: check this is this correct
<input type="checkbox" name="board[]" value="cbse" <?php if($val[0]=='cbse'){echo "checked";}?>>CBSE <input type="checkbox" name="board[]" value="cse/Ise" <?php if($val[1]=='ICSE/ISE'){echo "checked";}?>>ICSE/ISE.
解决方案You do the same as in your other question, but parse the JSON instead of exploding it.
$cats = json_decode($row['classconducted'], true); <input type="checkbox" name="segment[]" value="Class I-V Tuition" <?php if(in_array('Class I-V Tuition', $cats)) { echo "checked"; }?>> Class I-V Tuition
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