如何避免PHP的列索引？

Question

当我从一个表中获取用于创建Json格式化字符串的所有值时，我还在php response.How中获得了索引以避免它？

我的Php程序

<?php
include 'configa.php'; 
$sql1 = mysql_query("SELECT * FROM `message_store_push`") or die(mysql_error());
$c = mysql_num_rows($sql1);
$i=0;
while($sql2 = mysql_fetch_array($sql1)){
         $arrDet['AddressDetails']['details'][] = $sql2;
}

 if($c>0){
    $arrDet['Details']['status'] = "success";
 } else {
    $arrDet['Details']['status'] = "No data";
 }
echo json_encode($arrDet);
?>

而反应是

{
"AddressDetails":{
"details":[
{
"0":"1",
"key_id":"1",
"1":"2016-01-27 04:20:52",
"key_time":"2016-01-27 04:20:52",
"2":"HI",
"key_message":"HI"
}
]
},
"Details":{
"status":"success"
}
}

但我需要

{
"AddressDetails":{
"details":[
{
"key_id":"1",
"key_time":"2016-01-27 04:20:52",
"key_message":"HI"
}
]
},
"Details":{
"status":"success"
}
}

我怎样才能做到这一点

Answer 1

用这个：

<?php
include 'configa.php'; 
$sql1 = mysql_query("SELECT * FROM `message_store_push`") or die(mysql_error());
$c = mysql_num_rows($sql1);
$i=0;
while($sql2 = mysql_fetch_assoc($sql1)){ // instead of mysql_fetch_array()
         $arrDet['AddressDetails']['details'][] = $sql2;
}

 if($c>0){
    $arrDet['Details']['status'] = "success";
 } else {
    $arrDet['Details']['status'] = "No data";
 }
echo json_encode($arrDet);
?>

因为mysql_fetch_array()返回表值索引和关联键，所以只有关联键及其值总是使用mysql_fetch_assoc()函数！

Answer 2

您需要使用mysql_fetch_assoc而不是mysql_fetch_array

这

while($sql2 = mysql_fetch_array($sql1)){
         $arrDet['AddressDetails']['details'][] = $sql2;
}

应该是

while($sql2 = mysql_fetch_assoc($sql1)){
         $arrDet['AddressDetails']['details'][] = $sql2;
}

还请注意，mysql_*函数已从PHP7中删除，您可以开始使用mysqli或PDO。