我有一个名为 contracters_system 的表我想删除contracter_id中的特定值我该怎么做?
$plansarray = Array ( [0] => 4 [1] => 10 [2] => 47 );
$contractid = '21';
foreach ($plansarray as $servicesecondarray){
$servicequery = "select contracterid from contracters_system where planoption_id='$servicesecondarray'";
$servicequeryresult = Yii::app()->db->createCommand($servicequery)->queryAll();
$servicequery1 = "select contracterid from contracters_system where planoption_id <> '$servicesecondarray'";
$servicequeryresult1 = Yii::app()->db->createCommand($servicequery1)->queryAll();
$stack = json_decode($servicequeryresult[0]['contracterid'], true);
array_push($stack, $contractid);
if(empty($stack)){
$arraypush[] = $contractid;
} else {
$arraypush = $stack;
}
$finaldecode = json_encode(array_unique($arraypush));
if(!empty($finaldecode)){
$updatequery = "update contracters_system set contracterid = '$finaldecode' where planoption_id='$servicesecondarray' ";
$updateresult = Yii::app()->db->createCommand($updatequery)->query();
}
for($c=0;$c<count($servicequeryresult1);$c++){
$jsondecode = json_decode($servicequeryresult1[$c]['contracterid']);
//print_r($jsondecode);
if(($key = array_search($_GET['id'], json_decode($servicequeryresult1[$c]['contracterid'])) !== false) {
unset(json_decode($servicequeryresult1[$c]['contracterid'])[$key]);
// echo "test".$key;
$updatequery = "update contracters_system set contracterid = '".unset($jsondecode[$key])."' where planoption_id='$servicesecondarray' ";
$updateresult = Yii::app()->db->createCommand($updatequery)->query();
}
}
我在表格["1","21"]
中有 21 的价值。我的预期输出为["1"]
。
我如何在mysql和php中实现这一点?
答案 0 :(得分:0)
$value
。$decoded = json_decode($value)
$decoded
数组json_encode($updated_array)