json_en code在PHP [英] json_encode in php
本文介绍了json_en code在PHP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
我有这个脚本和它的作品,而不是如我所料。我需要用型动物的名称赋值的数组,现在都$常用3 []被命名为勇敢
i have this script and it works, but not as i expected. I need to assign an array of values with differents names, now all $arr[] are named "valor"
{"valor":"20"},{"valor":"50"}
我需要
{"valor1":"20"},{"valor2":"50"}
剧本
$query = mysql_query("SELECT valor FROM grafico") or die(mysql_error());
$arr = array();
while($row = mysql_fetch_assoc($query)) {
$arr[] = $row;
}
echo json_encode($arr);
in ajax
<script type="text/javascript">
jQuery(document).ready(function(){
jQuery("button").click(function(){
jQuery.ajax({
url: "chart.php",
dataType: "json",
success: function(json){
var msg = "Nome: " + json.valor1+ "\n";
msg += "Sobrenome: " + json.valor2 + "\n";
alert(msg);
}
});
});
});
</script>
问题是:我需要创造,创造独有名称的循环,因为值1,值2,值3
the problem is: I need to create a loop that create uniques names, as value1, value2, value3
喜欢
$改编['valor1'] =伊万德罗;
我尝试了LOOP-的,但我得到的内存错误
i tried a loop- for, but i get an error of memory
感谢
推荐答案
试试这个:
$query = mysql_query("SELECT valor FROM grafico") or die(mysql_error());
$arr = array();
$i = 1;
while($row = mysql_fetch_assoc($query)) {
$arr[] = array("valor{$i}" => $row["valor"]);
++$i;
}
echo json_encode($arr);
应该工作。另外,如果你想它,所以它的工作原理与目前的回调修改 $ ARR [] =
线以下内容:
$arr["valor{$i}"] = $row["valor"];
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