我的问题是,我有一个SQL查询填充数组,并希望将数组合并到一个新的数组中,其中查询的ID不是重复的,我试过了merge_array和当我通过URL执行函数时,它不显示数组中的内容我只是得到ARRAY.Have也试图将数组连接到另一个数组并获得相同的结果,再次我的问题是如何加入数组并正确显示。PHP将数组连接到数组上sql查询的结果
$rows[0] = array("181", "a","g"); //Results from previous query
$rows[1] = array("181","j","L")
$rows[2] = array("181","p");
$rows[3] = array("182","k");
$rows[4] = array("183","l");
$rows[5] = array("183","p");
$id =0;
$commentsH = "";
while($row=mysql_fetch_array($query_comments)){
If($id == $image){ //image id is the first element in array.
$comments[] =$row;
$commentsH = $comments.",".$commentsH[$i];
}
else{
$id = $image;
$i = $i +1;
}
}
$result = array();
$result["result"] = 500;
$result["message"] = "Database Read Successfully";
$result["comments"] =$commentsH;
echo json_encode($result);
exit;
EXPECTED OUTPUT
$commentsH[0] = array("181", "a","g","j","L","p");
$commentsH[1] = array("182","k");
$commentsH[2] = array("183","l","p");
那么...声明'$ image'在哪里?您在每次迭代和代码中覆盖'$ commentSH','$ commentsH'不是您的_expected output_中显示的数组,'$ commentsH = $ comments。“,”。$ commentsH [$我];'没有任何意义.. – dbf 2013-04-28 10:02:35
..有什么你需要知道关于使用[mysql函数](http://php.net/manual/en/function.mysql-query.php ) – dbf 2013-04-28 10:07:51
$ image是一个查询的结果,我没有在上面认为这很明显,$ commentsH = $ comments。“,”。comments [$ i]在两个数组之间用“,”连接数组如果陈述是真的,在其他语言C和Java中很常见。 – Freddy 2013-04-28 10:13:32