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我有這個實現與融合表層，我嘗試使用字母A-Z的預定義標記圖標來顯示搜索查詢在地圖上的結果（很像原來的谷歌地圖）。融合層中可以有多少個不同的標記是否有限制？
我實現這一目標是先建立一個層，與所有標記的通用圖標的方式..
var layer = new google.maps.FusionTablesLayer({
query: {
select: 'Geometry',
from: 0000000
},
map: map,
options: {
suppressInfoWindows: true
},
styles: [{
markerOptions: {
iconName: "measle_white"
}
}]
});
同時，我詢問同桌的25個結果與ST_DISTANCE排序基於我的地方地圖的中心（geopositioned）
var queryUrlHead = 'http://www.google.com/fusiontables/api/query?sql=',
queryUrlTail = '&jsonCallback=success',
query = 'SELECT+ID+FROM+0000000+ORDER+BY+ST_DISTANCE(Geometry,LATLNG('+position.coords.latitude+','+position.coords.longitude+'))+LIMIT+25';
var queryurl = queryUrlHead + query + queryUrlTail;
返回的JSON對象是唯一的ID的我稱之爲「IDS」的數組。然後我使用一些觸發器（zoomchanged）來重新繪製帶有字母圖標的25個最近的圖標（由this啓發）。
google.maps.event.addListener(map, 'zoom_changed', function() {
layer.setOptions({
styles: [{
markerOptions: {
iconName: "measle_white"
}
}, //fallback
{
where: "'ID' = " + ids.table.rows[0][0],
markerOptions: {
iconName: "a_blue"
}
}, {
where: "'ID' = " + ids.table.rows[1][0],
markerOptions: {
iconName: "b_blue"
}
}, {
where: "'ID' = " + ids.table.rows[2][0],
markerOptions: {
iconName: "c_blue"
}
}, {
where: "'ID' = " + ids.table.rows[3][0],
markerOptions: {
iconName: "d_blue"
}
}, {
where: "'ID' = " + ids.table.rows[4][0],
markerOptions: {
iconName: "e_blue"
}
}, {
where: "'ID' = " + ids.table.rows[5][0],
markerOptions: {
iconName: "f_blue"
}
}, {
where: "'ID' = " + ids.table.rows[6][0],
markerOptions: {
iconName: "g_blue"
}
}, {
where: "'ID' = " + ids.table.rows[7][0],
markerOptions: {
iconName: "h_blue"
}
}, {
where: "'ID' = " + ids.table.rows[8][0],
markerOptions: {
iconName: "i_blue"
}
}, {
where: "'ID' = " + ids.table.rows[9][0],
markerOptions: {
iconName: "j_blue"
}
}, {
where: "'ID' = " + ids.table.rows[10][0],
markerOptions: {
iconName: "k_blue"
}
}, {
where: "'ID' = " + ids.table.rows[11][0],
markerOptions: {
iconName: "l_blue"
}
}, {
where: "'ID' = " + ids.table.rows[12][0],
markerOptions: {
iconName: "m_blue"
}
}, {
where: "'ID' = " + ids.table.rows[13][0],
markerOptions: {
iconName: "n_blue"
}
}, {
where: "'ID' = " + ids.table.rows[14][0],
markerOptions: {
iconName: "o_blue"
}
}, {
where: "'ID' = " + ids.table.rows[15][0],
markerOptions: {
iconName: "p_blue"
}
}, {
where: "'ID' = " + ids.table.rows[16][0],
markerOptions: {
iconName: "q_blue"
}
}, {
where: "'ID' = " + ids.table.rows[17][0],
markerOptions: {
iconName: "r_blue"
}
}, {
where: "'ID' = " + ids.table.rows[18][0],
markerOptions: {
iconName: "s_blue"
}
}, {
where: "'ID' = " + ids.table.rows[19][0],
markerOptions: {
iconName: "t_blue"
}
}, {
where: "'ID' = " + ids.table.rows[20][0],
markerOptions: {
iconName: "u_blue"
}
}, {
where: "'ID' = " + ids.table.rows[21][0],
markerOptions: {
iconName: "v_blue"
}
}, {
where: "'ID' = " + ids.table.rows[22][0],
markerOptions: {
iconName: "w_blue"
}
}, {
where: "'ID' = " + ids.table.rows[23][0],
markerOptions: {
iconName: "x_blue"
}
}, {
where: "'ID' = " + ids.table.rows[24][0],
markerOptions: {
iconName: "z_blue"
}
}
]
});
現在，除了前5個結果A-D（回退圖標的+1）外，這個實際上效果非常出色。這裏出了什麼問題？我是否遇到了一些限制（markerOptions只有5個值）還是我弄亂了代碼？
旁註：This example每層看起來有5個以上的圖標，但Google做了它，我不明白它中的任何一個。