我如何获得字符串输出?我尝试解码时发生了一个错误。我想在文本框中插入输出值。
怎么做到呢 ?
$array=json_decode($json);
echo $array;
**Warning:**
json_decode() expects parameter 1 to be string, array given in C:\xampp\htdocs\school\vijay\update.php on line 20
我的PHP
<?php
$json = array();
$con=mysql_connect("localhost","school","certify");
$db_select = mysql_select_db('School_Data', $con);
$childid = $_GET['childid'];
$result = mysql_query("SELECT * FROM childinfo where ChildID='$childid'",$con);
while($r = mysql_fetch_assoc($result)) {
$json[] = $r;
}
if($result){
echo json_encode($json);
}
else
{
echo mysql_error();
}
//$obj = unserialize($json);
$arrayOfEmails=json_decode($json);
echo $arrayOfEmails;
mysql_close($con);
?>
我的JSON输出
[{
"ID": "1",
"ChildID": "1001",
"ParentID": "2002",
"SiblingsID": "hfh",
"TeacherID": "hfhf",
"ChildName": "fhfh",
"DOB": "2014-03-04",
"Age": "0",
"Gender": "male",
"Grade": "KG1",
"Section": "KG1",
"Stream": "NORMAL",
"BloodGroup": "O-",
"Nationality": "KG1",
"Country": "Lebanon",
"Religion": "KG1",
"MotherTongue": "KG1",
"FirstLanguage": "bfbf",
"SecondLanguage": "fbfbfb",
"PlaceOfBirth": "fhfh",
"LandlineNumber": "0",
"EmailID": "[email protected]",
"ChildPhoto": "Requirement.PNG",
"TemporaryAddress": "bfdbd",
"PermanentAddress": "bdbdbf",
"Mentor": "fbbfd",
"DateOfJoin": "2014-03-06",
"JoinGrade": "J",
"ReferredBy": "bdbf",
"ContactNumber": "0",
"EmergencyContactNumber": "0"
}]
保存json编码的数组,然后稍后将其还原。
未经测试的代码
<?php
$json = array();
$con=mysql_connect("localhost","school","certify");
$db_select = mysql_select_db('School_Data', $con);
$childid = $_GET['childid'];
$result = mysql_query("SELECT * FROM childinfo where ChildID='$childid'",$con);
while($r = mysql_fetch_assoc($result)) {
$json[] = $r;
}
if($result){
$encodedJson = json_encode($json); // save json string for later
echo $encodedJson;
}
else
{
echo mysql_error();
}
//$obj = unserialize($json);
$arrayOfEmails = json_decode($encodedJson); // convert back to an array
var_dump($arrayOfEmails);
mysql_close($con);
?>
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